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y x y x
, ,
,
.
r r x y
Let P′ ( x′, y′ ) be another point on the terminal side of α .
Chapter 15 Solving Triangles
I. Trigonometric functions
15.1
Trigonometric functions
In the construction of a water
P′
supply system, pipes are laid along a
P
slope which makes an angle α with
h
the horizontal. As in diagram 15-1, the
pipe starts from the bottom of slope O
α
to point P. The height of P from the O
M M′
horizon is PM. Suppose the pipe
Diagram 15-1
extends further up to P′ , with M ′P ′
the height of P′ , then the longer the pipe gets, the higher is the level
of the top of the pipe from the horizon. It is easy to see that △OPM
MP M ′P′
～△ OP ′M ′ , and therefore
. That is to say, when
=
OP OP′
angle α is fixed, the ratio of the height of the top of the pipe to the
length of the pipe is a fixed value.
Think for a while: What is the above ratio when α is 30°?
This example inspires us to study further the ratios of the sides
relating to an angle of a triangle.
Suppose α is an angle, we
y
align the vertex of the angle α
with the origin and the initial side of
P′( x′, y′)
the angle α with the positive axis
P(x, y)
Ox. A rectangular coordinate system
r
can be built as in diagram 15-2. Take
y
any point P(x, y) on the terminal side
α
x
of angle α . Its distance to the origin O
x
′
O(0,0) is r = x 2 + y 2 (r is always
positive). We have a set of ratios:
- 256 -
M
Diagram 15-2
M
Distance from P′ to the origin O(0, 0) is r ′ = x′2 + y ′2 . We have
another set of ratios:
y′ x′ y′ x′
.
,
,
,
r ′ r ′ x′ y′
Through P and P′ draw lines MP and M ′P ′ perpendicular to the
x-axis, then △OPM～△ OP ′M ′ . The signs of x and x′ , y and y′ are
the same, hence
y y ′ x x′ y y ′ x x ′
= .
= ,
= ,
= ,
r r ′ r r ′ x x′ y y ′
From this we know that given any fixed angle α ，these four ratios are
determined only by the size of α , and are independent of the
position of P on the terminal side of angle α . Therefore these ratios
are functions of α . We call
y
y
the sine of α , denoted by sin α , i.e. sin α = ;
r
r
x
x
the cosine of α , denoted by cos α , i.e. cos α = ;
r
r
y
y
the tangent of α , denoted by tan α , i.e. tan α = ;
x
x
x
x
the cotangent of α , denoted by cot α , i.e. cot α = .
y
y
sin α , cos α , tan α , cot α are called trigonometric functions of α .
Note: sin α is a whole sign, it stands for the sine value of α and
should not be understood as sin i α . This is also true for the
other trigonometric functions of α .
【Example 1】Given that the terminal side of α passes through
P(3, 4), find the four trigonometric functional values
of α (diagram. 15-3).
- 257 -
Note: We may also call these values the trigonometric values of α .
Solution ∵
∴
∴
r = x + y = 3 +4 =5
y 4
sin α = = ,
r 5
x 3
cos α = = ,
r 5
y 4
tan α = = ,
x 3
x 3
cot α = = .
y 4
2
2
2
2
y
P (3, 4)
α
x
O
Diagram 15-3
【Example 2】Prove that
(1) tan α =
Proof
sin α
;
cos α
(2) sin 2 α + cos 2 α = 1 .3
(1) From the definitions of trigonometric functions,
y
sin α r y
= = = tan α
cos α x x
r
that is,
sin α
tan α =
.
cos α
2
2
y2 + x2
 y x
(2) sin 2 α + cos 2 α =   +   =
.
r2
r r
∵ y2 + x2 = r 2
∴
sin 2 α + cos 2 α =
Practice
1. (Mental) What kind of trigonometric function of angle β is
sin β ? How is it expressed as a ratio? What about tan β ,
cos β , cot β ?
2. Given that the terminal side of angle α passes through the point
in the four cases below, find the values of the four trigonometric
functions of α for each case.
(1) (4, 3)； (2) (5, 12)； (3) (2, 2)； (4) (2, 3)。
x = 3, y = 4
r2
=1
r2
sin 2 α stands for (sin α ) 2 , the same convention of raising a trigonometric
function to a certain power is adopted for all other trigonometric functions.
15.2
Values of trigonometric functions for 30°, 45°, 60°
We know that given any angle α , its four trigonometric values
are uniquely determined. For some special angles, we can calculate
the values as follows.
(1) In diagram 15-4, α = 30° . We pick a point P on the terminal
side of angle α . Let the y-coordinate of P be a. Through P draw a
perpendicular line to the x-axis MP. In the right-angled triangle OPM,
∠POM = 30°, MP = a, and OP = 2a. (Why?) From Pythagoras’
Theorem,
OM = (2a ) 2 − a 2 = 3a ,
That is to say, the coordinates of P are ( 3a , a ), r = 2 a . Therefore
y a 1
sin 30° = =
=
y
r 2a 2
P
x
3a
3
a
cos 30° = =
=
r
2a
2
2a
y
a
3
tan 30° = =
=
30°
x
x
3
3a
O
M
3a
cot 30° =
x
3a
=
= 3
y
a
3
- 258 -
- 259 -
Diagram 15-4
(2) In diagram 15-5, α = 45° . We take a point P on the terminal
side of angle α . Let the y-coordinate of P be a, then P has
x-coordinate a. (Why?) From Pythagoras’ Theorem, r = 2a .
Therefore
2
y
a
sin 45° = =
=
y
r
2
2a
x
=
r
y
tan 45° = =
x
x
cot 45° = =
y
cos 45° =
a
45°
Sine
1
2
2
2
3
2
Cosine
3
2
2
2
1
2
Tangent
3
3
1
3
Cotangent
3
1
3
3
x
a
M
Diagram 15-5
(3) In diagram 15-6, α = 60° . We take a point P on the terminal
side of angle α . Let the x-coordinate of P be a, then r = OP = 2a .
(Why?) From Pythagoras’ Theorem, P has y-coordinate
y = (2a ) 2 − a 2 = 3a . Therefore
sin 60° =
y
P
2a
O
3a
60°
a M
Diagram 15-6
【Example】 Evaluate
(1) 2 sin 30° + 3cos 60° + tan 45° ;
(2) sin 2 45° + cot 60° cos 30° ;
1
2
(3)
cos 30° +
cos 45° + sin 60° cos 60° .
2
2
1
1
1
Solution (1) 2 sin 30° + 3cos 60° + tan 45° = 2 × + 3 × + 1 = 3 ;
2
2
2
2
x
We will be using the trigonometric values above quite often. For
ease of remembering, they are listed in the table below:
- 260 -
60°
2a
O
y
3a
3
=
=
r
2a
2
x a 1
cos 60° = =
=
r 2a 2
y
3a
tan 60° = =
= 3
x
a
3
x
a
cot 60° = =
=
3
y
3a
45°
P
2
a
=
2
2a
a
=1
a
a
=1
a
30°
3
3


(2) sin 2 45° + cot 60° cos 30° =  2  +
×
3
2
 2 
1 1
= + =1
2 2
1
2
(3)
cos 30° +
cos 45° + sin 60° cos 60°
2
2
1
3
2
2
3 1 2+2 3
= ×
+
×
+
× =
2 2
2
2
2 2
4
1+ 3
=
2
- 261 -
Practice
1. (Mental) What are the values of sin 30° and cos 60° ? tan 45°
and cot 45° ? sin 45° and cos 45° ? sin 60° and cos 30° ?
tan 60° and cot 30° ? tan 30° and cot 60° ?
2. Find the value of the following expression.
(1) sin 30° − 3 tan 30° + 2 cos 30° ;
(2) 2 cos 30° + tan 60° − 6 cot 60° ;
(3) 5cot 30° − 2 cos 60° + 2sin 60° ;
(4) cos 2 45° + sin 2 45° ;
sin 60° − cot 45°
(5)
.
tan 60° − 2 tan 45°
3. On a rectangular coordinate system, use the origin as the vertex,
positive x-axis as the initial side, draw an angle of 40° in the
first quadrant. Measure the distance of a point on the terminal
side to the origin, then calculate the four trigonometric values of
40° with accuracy to 0.01.
15.3 Using a scientific calculator to find trigonometric
values* (This section can be skipped)
In the last section we have found the trigonometric values for
30°, 45°, 60°, but for an arbitrary angle this would be a problem. In
the past, this was done by looking up the values from trigonometric
tables. Now, we can get much more accurate values by use of a
calculator.
All windows operation systems come with the little application
“Calculator”. On starting the application under the view menu, there
is the choice of “Scientific mode”. Students may also search the
internet for other on-line calculators to use.
Each of these “calculators” may have its own function keys
which are different from other calculators:
(i) The 2ndF button in this section is the secondary function button,
(ii) In some calculators it is the SHIFT button, and
(iii) In the Windows Scientific Calculator, it is the INV button.
- 262 -
The buttons DEG and DMS are function keys for converting
angle measurement in between two formats, namely the degree in
decimal format and the degree in Degree-Minute-Second format.
When using the DMS button, there may be three sets of numbers.
The first set shows the Degree, the second set shows the Minutes and
the third set shows the Second.
1.
Sine and Cosine
【Example 1】Use calculator to find sin10°36′ (accurate to 4
decimal places).
Solution On a scientific calculator, press successively
sin , □
1 , □
0 , DMS , □
3 , □
6 , DMS , □
=
The display shows 0.18395135…, the answer is therefore
0.1840.
If the Windows Calculator is used, press
1 , □
0 , □
˙, □
3 , □
6 ,
□
Inv , deg , sin
The display shows 0.18395135…, the answer is 0.1840.
Answer: 0.1840.
Note：In the Windows Calculator, the angle is entered before the
function.
Practice
Use a scientific calculator to find (accurate to 4 decimal places):
(1) sin 28°30′ ；
(2) cos 35′ ；
(4) cos 62°48′ 。
(3) sin 60°48′ ；
【Example 2】Given that sin α = 0.3688 , use a scientific calculator
to find α (accurate to 1′ ).
Solution Press successively
2ndF , sin , □
0 , □
˙, □
3 , □
6 , □
8 , □
8 , □
=
The display shows 21.6416292… (denary expression),
then press
2ndF , DMS
- 263 -
The display changes to 21°38′29.87′′ , i.e. α ≈ 21°38′ .
If the Windows Calculator is used, press
0 , □
˙, □
3 , □
6 , □
8 , □
8 ,
□
Inv , sin-1
The display shows 21.6416292… (denary expression),
then press
dms
then the display changes to 21.382986511…,
i.e. α = 21°38′29.86511⋯′′ ≈ 21°38′ .
Answer: 21°38′ .
Note：In using scientific calculator to find inverse trigonometric
functional values, the value is entered before pressing the
function button.
Practice
Use a scientific calculator to find the acute angle α (accurate to
1′ ):
(2) sin α = 0.1436 , α = ?
(1) sin α = 0.8268 , α = ?
(3) cos α = 0.3279 , α = ?
(4) cos α = 0.9356 , α = ?
2.
Tangent and Cotangent
The procedures in finding the tangent and cotangent and their
inverse functions are similar to the calculation for sine and cosine.
We shall omit the description.
【Example 1】Use calculator to find tan 53°49′ (accurate to 4
decimal places).
Solution On a scientific calculator, press successively
tan , □
5 , □
3 , DMS , □
4 , □
9 , DMS , □
=
If the Windows Calculator is used, press
5 , □
3 , □
˙, □
4 , □
9 ,
□
Inv , deg ,
tan
The display shows 1.36716099…, the answer is 1.3672.
Answer: 1.3672.
- 264 -
【Example 2】Use calculator to find cot14°32′ (accurate to 4
decimal places).
Solution On a scientific calculator, press successively
1 , □
÷ , tan , □
1 , □
4 , DMS , □
3 , □
2 , DMS ,
□
= .
□
The display shows 3.85745373…, the answer is
3.8575.
If the Windows Calculator is used, press
1 , □
/ , □
1 , □
4 , □
˙, □
3 , □
2 , Inv , deg ,
□
tan , □
= . The display shows 3.85745373…, the answer
is 3.8575.
Answer: 3.8575.
Practice
Use a scientific calculator to find (accurate to 4 decimal places):
(1) tan13°12′ ; (2) tan 40°55′ ; (3) tan 54°28′ ; (4) tan 89°43′ ;
(5) cot 72°18′ ; (6) cot 56°56′ ; (7) cot 32°32′ ; (8) cot15°15′ .
【Example 3】Given that tan α = 1.4036 , use a scientific calculator
to find α (accurate to 1′ ).
Solution Press successively
2ndF , tan , □
1 , □
˙, □
4 , □
0 , □
3 , □
6 , □
=
The display shows 54.5318877… (denary expression), then
press
2ndF , DMS
The display changes to 54°31′ 54′′ , i.e. α ≈ 54°32′ .
If the Windows Calculator is used, press
1 , □
˙, □
4 , □
0 , □
3 , □
6 ,
□
Inv , tan-1 , □
=
The display shows 54.53188… (denary expression),
then press
dms
- 265 -
then the display changes to 54.31547960…,
i.e. α ≈ 54°32′ .
Answer: 54°32′ .
【Example 4】Given that cot α = 0.8637 , use a scientific calculator
to find α (accurate to 1′ ).
Solution Press successively
2ndF , tan , □
1 , □
÷ , □
0 , □
˙, □
8 , □
6 , □
3 ,
7 , □
= .
□
The display shows 49.18282777… (denary
expression), then press
2ndF , DMS
The display changes to 49°10′58′′ , i.e. α ≈ 49°11′ .
If the Windows Calculator is used, press
1 , □
/ , □
0 , □
˙, □
8 , □
6 , □
3 , □
7 , □
= ,
□
Inv ,
-1
tan . The display shows 49.1828277 … (denary
expression), then press
dms
The display changes to 49.10581799…, i.e. α ≈ 49°11′ .
Answer: 49°11′ .
Practice
Use a scientific calculator to find the acute angle α (accurate to
1′ ):
(1) tan α = 0.9131 , α = ?
(2) tan α = 0.3314 , α = ?
(3) tan α = 2.220 , α = ?
(4) tan α = 31.80 , α = ?
(5) cot α = 1.6003 , α = ?
(6) cot α = 3.590 , α = ?
(7) cot α = 0.0781 , α = ?
(8) cot α = 180.9 , α = ？
- 266 -
Exercise 9
1. Given that the terminal side of α passes through the points in
each of the four cases below，find the four trigonometric values of
angle α in each case.
(1) (1, 2);
(3) (2, 5); (4) ( 2 , 3 ).
(2) (1, 2 );
2. Evaluate
(1) 3 tan 30° + cot 45° − 2 tan 45° + 2sin 60° ;
(2) 2 cos 30° − 4 cot 30° + 3 tan 60° ;
cos 45° − sin 30°
(3)
;
(4) sin 2 45° + cos 2 60° ;
cos 45° + sin 30°
3cot 60°
cos 60°
1
(5)
;
(6)
+
.
2
2 cos 30° − 1
1 + sin 60° tan 30°
3. Use a scientific calculator to find
(1) sin 28°18′ , sin 57°43′ , sin 68°33′ , sin 72°58′ ;
(2) cos 65°2′ , cos10°36′ , cos 44°15′ , cos 32°4′ .
4. Answer the following:
(1) Is sin 20° + sin 40° equal to sin 60° ?
(2) Is cos10° + cos 20° equal to cos 30° ?
5. Find the acute angle for each equation.
(1) sin α = 0.6841 , cos α = 0.3241 , sin α = 0.5136 ;
(2) cos β = 0.2839 sin β = 0.0526 , cos β = 0.5412 .
6. Evaluate:
(1) tan 9°19′ , tan 64°10′ , tan 75°39′ , tan 79°51′ ;
(2) cot 8°28′ , cot16°25′ , cot 48°27′ , cot 88°44′ ;
(3) sin 89°32′ , cos 38°43′ , tan 5′ , cot14°27′ .
7. Find the acute angle for each equation
(1) tan α = 0.7817 , cot α = 1.1106 , tan α = 1.0736 ；
(2) cot β = 3.267 , tan β = 2.378 , cot β = 57.82 ;
2
(3) sin x = 0.86276 , cos x = , tan x = 53.10 .
3
- 267 -
II. Solving right-angled triangles
15.4 Relations between sides and angles in a
right-angled triangle
y
y
If α represents an acute angle of
a right-angled triangle, then (1) and (2)
can be described as (Diagram 15-9)
opposite
side
In production work and in scientific research, it is frequently
required to measure or calculate the lengths and angles of a geometry
shape. The problem usually can be reduced to finding the sides or
angles of a triangle. Calculate the unknown sides and angles from the
known sides and angles of a triangle is called solving a triangle. We
begin with solving right-angled triangles. For this, we use
trigonometric functions to relate the sides and angles of a right-angle
triangle.
As in diagram 15-7, in right-angled
triangle ABC, C is the right angle and
B
the hypotenuse is c. Acute angle A has
c
a
opposite side a and adjacent side b.
Acute angle B has opposite side b and A
C
b
adjacent side a.4 As in diagram 15-8 (1)
Diagram 15-7
or (2), a rectangular coordinate system is
built.
From the definitions of trigonometric functions we get
a
b
a
b
(1) sin A = , cos A = , tan A = , cot A = ;
c
c
b
a
b
a
b
a
(2) sin B = , cos B = , tan B = , cot B = .
c
c
a
b
hypotenuse
α
adjacent side
Diagram 15-9
Opposite side of α
Adjacent side of α
, cos α =
Hypotenuse
Hypotenuse
Opposite side of α
Adjacent side of α
tan α =
, cot α =
Adjacent side of α
Opposite side of α
sin α =
These four formulas give the relations between the sides and the
angles of a right-angled triangle. From now on, when we solve
right-angled triangles there is no need to use the coordinate system
and the formulas can be applied directly.
Since B = 90° − A , From (1) and (2) we also get
sin(90° − A) = cos A , cos(90° − A) = sin A
A (a, b)
tan(90° − A) = cot A , cot(90° − A) = tan A
c
B (b, a)
c
O A
b
(1)
b
a
C
x
O B
a
C
(2)
x
Diagram 15-8
4
In this chapter, we adopted the convention of using a, b, c to represent the sides
and A, B, C to represent the angles of a right-angled triangle.
- 268 -
Since the trigonometric functions of 90° − A and
A are related this way, sine and cosine can share
the same mathematics table when we need to look
up their values from mathematics table. The same
is true for tangent and cotangent.
【Example】 In the right-angled triangle ABC,
given a = 12 , b = 5 , find the four
trigonometric values for angle A
and for angle B (Diagram 15-10).
- 269 -
B
c
a
A
C
b
Diagram 15-10
Solution From Pythagoras’ theorem, we get
15.5 Solving right-angled triangles
c = a + b = 12 + 5 = 13
a 12
b 5
∴ sin A = = , cos A = = ,
c 13
c 13
a 12
b 5
tan A = = , cot A = = ,
b 5
a 12
5
12
sin B = cos A = , cos B = sin A = ,
13
13
5
12
tan B = cot A = , cot B = tan A = .
12
5
2
2
2
2
Practice
1. (Mental) Find the four trigonometric values in each of the
diagrams:
C
B
We know that for a right-angled triangle, there are 6 elements: 3
sides and 3 angles. Excluding the right angle, the relations between
the elements are
(1) 3 sides:
a 2 + b 2 = c 2 (The Pythagoras’ Theorem)
(2) Two acute angles:
A + B = 90°
(3) Between sides and angles
Opposite site of α
Adjacent site of α
sin α =
, cos α =
Hypottenuse
Hypottenuse
Opposite site of α
Adjacent site of α
tan α =
, cot α =
Adjacent site of α
Opposite site of α
In the formula, the symbol α represents the actue angle A
or acute angle B.
5
If we know any two elements other than the right-angle (one of
the two elements must be a side) of a right-angled triangle, then we
can use one of the above formula to calculate a third element, and
hence calculate all the 3 unknown elements.
2 14
10
6
A
C
8
9
A
(No. 1)
B
2. In a right-angled triangle,
(1) Given that a = 2 , b = 1 , find the four trigonometric values
of angle A.
(2) Given a = 3 , b = 4 , find the four trigonometric values of
angle B.
(3) Given b = 2 , c = 29 , find the four trigonometric values
of angle A and angle B.
3. Write the following expressions as trigonometric functions of
angles A or B.
(1) cos(90° − A) ;
(2) tan(90° − B ) ;
(3) sin(90° − B ) ;
(4) cot(90° − A) .
- 270 -
【Example 1】In diagram 15-11, ABC is a right-angled triangle.
Given that b = 35 , c = 45 , find A, B (accurate to 1° )
and a (correct to 2 significant figures).
b 35
Solution (1) cos A = =
≈ 0.7778 ,
B
c 45
A ≈ 39° using calculator..
(2) B = 90° − A ≈ 90° − 39° = 51° .
45
(3) a = c 2 − b 2
α
= 452 − 352
= 800
a = 28.28 ≈ 28 using calculator.
- 271 -
A
C
35
Diagram 15-11
In example 1, since sides b, c are known; to find angle A, we
choose an equation that includes angle A and sides b, c, which is
b
cos A = .
c
【Example 2】In right-angled triangle ABC, given that a = 15 ,
A = 35°27′ , find b, c and B (angle accurate to 1° ,
sides correct to 2 significant figures).
b
Solution (1) ∵ cot A =
a
∴ b = a cot A = 15 × cot 35°27′ = 15 × 1.4045 ≈ 21
a
(2) ∵ sin A =
c
a
15
15
∴ c=
=
=
≈ 26
sin A sin 35°27′ 0.5800
(3) B = 90° − 35°27′ = 54°33′
In example 2, a and A are known. To find b, we can choose
a
b
between two functions, namely (i) tan A = , and (ii) cot A = . In
b
a
a
case (i), b =
, where division by tan A is difficult to calculate.
tan A
In case (ii), b = a cot A , where multiplication by cot A is easier to
calculate. Therefore it is recommended to use the second formula.
Practice
1. In the right-angled triangle ABC,
(1) Given c, A, write down the relation between a and b.
(2) Given b, A, write the equation to find a. Given a, A, write
the equation to find b.
(3) How to find A given a and b ? How to find A given a and
c ? How to find A given b and c ?
- 272 -
Practice
2. Solve the right-angled triangle given the conditions.
(1) c = 10 , A = 30° ;
(2) b = 15 , B = 45° ;
(3) a = 51 , c = 70 ;
(4) a = 22 , b = 12 .
(Correct to 1° for (3) and (4), correct to 2 significant figures for
the sides)
15.6 Examples of application
There are wide applications of solving right-angled triangles.
We give some examples below.
【Example 1】The roof of a factory is supported by rafters and a
main pillar in the centre (equilateral triangle). The
span of the structure is 10 m and angle A
is 26° (diagram 15-12). Find the length of the main
pillar BC (C is the midpoint of the base) and the
principal rafter AB (accurate to 0.01 m).
B
rafter
pillar
A
26°
C
span
Diagram 15-12
Solution △ ABC is a right-angled triangle, with ∠C = 90° ,
∠A = 26° and AC = 5 m.
BC
∵ tan A =
AC
∴ BC = AC i tan A = 5 × tan 26° = 5 × 0.4877 ≈ 2.44 m
AC
∵ cos A =
AB
AC
5
∴ AB =
=
cos A cos 26°
- 273 -
5
≈ 5.56 m。
0.8988
Answer: BC is 2.44 m, AB is 5.56 m.
By calculator, AB ≈
【Example 2】 The cross-section of a dovetail groove is an isosceles
trapezium. Diagram 15-13 is one such cross-section,
with the dovetail angle ∠B equal to 55° , and width
of the upper mouth AD equal to 180 mm. Depth of
the dovetail groove is 70 mm. Find the width of the
base BC (accurate to 1 mm).
B
A
D
E
F
C
Diagram 15-13
Analysis: Connect AD and draw AE, DF perpendicular to BC, then
AD = EF, BE = FC. Also BC = BE + EF + FC = 2 BE
+ AD . Since AD is known, we only have to find BE to
solve the problem.
Solution Draw AE ⊥ BC , DF ⊥ BC . In the right-angled triangle
△ABE,
BE
∵ cot B =
AE
∴ BE = AE i cot B = 70 × cot 55° = 70 × 0.7002 ≈ 49.0 mm
∴ BC = 2 BE + AD ≈ 2 × 49.0 + 180 = 278 mm
Answer:The base width of the dovetail groove is approximately
278 mm.
【Example 3】As in diagram 15-14, trees are planted along a hill
slope. It is required that the trees should be 5.5 m
apart in horizontal distance. The inclination of the
slope is measured to be 24° . find the distance along
the slope between two adjacent trees. (accurate to
1 mm).
- 274 -
B
C
(1)
24°
5.5 m
(2)
A
Diagram 15-14
Analysis: As in diagram 15-14(2), ∠A = 24° . Horizontal distance
AC = 5.5 m. BC ⊥ AC . △ABC is a right-angled triangle.
The problem is solved by finding the length of AB from
△ABC.
Solution In △ABC, ∠C = 90°
AC
∵ cos A =
AB
AC
5.5
∴ AB =
=
≈ 6.0 m
cos A 0.9135
Answer: The distance between two adjacent trees is 6.0 m.
Whether in building a dam or a
water trench, widening a river or in
road construction, the engineering
i = h：l
h
specification must label the steepness
of slopes which is called the gradient.
The gradient is defined to be the ratio
l
of the height h of the slope to its
Diagram 15-15
horizontal width l . (Diadram 15-15).
If i is the gradient,
h
i= .
l
The gradient is often written in the form 1：m, such as i = 1：5
1
( i = ).
5
- 275 -
If α is the angle of slope, then
h
i = = tan α .
l
Obviously if the gradient gets larger (and therefore α ), the slope
gets steeper.
BE
BE
=1：3, we can also get
=1： 10 . .
AE
AB
Therefore AB = 10 i BE = 10 × 23 ≈ 72.7 m.
【Example 4】The cross-section of a dam is a trapezium (diagram
15-16). The width of the top of the dam is 6 m. The
height of the dam is 23 m. Slope AB has gradient i =
1：3. Slope CD has gradient i′ = 1：2.5. Find the
angle of slope α of AB, the width of the dam AD,
and the length of slope AB (accurate to 0.01 m).
1. As in diagram, a factory building has a supporting structure in
shape of an isosceles triangle with base AB = 12 m. Find the
length of the central pillar CD and the rafter AC. (accurate to 0.1
m)
In example 4, from
Practice
C
C
C
B
5m
6
1：2.5
1：3
A
23
A
α
E F
Diagram 15-16
D
Solution In diagram 15-16, draw BE ⊥ AD and CF ⊥ AD . In the
right-angled triangles ABE and CDF,
BE
CF
∵
= 1：3,
= 1：2.5
AE
FD
∴ AE = 3BE = 3 × 23 = 69 m
FD = 2.5CF = 2.5 × 23 = 57.5 m
∴ AD = AE + EF + FD = 69 + 6 + 57.5 = 132.5 m
1
The gradient of AB is i = tan α = , from which we obtain
3
α ≈ 18°26′ using a scientific calculator.
BE
∵
= sin α
AB
BE
23
∴ AB =
=
= 72.73 ≈ 72.7 m
sin α sin18°26′
Answer: Angle of slope AB α is 18°26′ . Dam width AD is 132.5 m.
Length of AB is 72.7 m.
- 276 -
A
B
D
D
(No. 1)
B
(No. 2)
2. As in diagram, wires are drawn from an electricity post at a
point C 5m above the ground to stabilize it. The wires make an
angle of 60° with the horizontal. Find the length of wire AC
and the distance of A from the bottom of the pole D. (accurate to
0.01 m)。
3. As in diagram, the hill is drilled along the direction AC for pipe
construction. To speed up, drilling is to proceed simultaneously
on the other side of the hill at E. Pick a point B on AC so
that ∠ABD = 140° , BD = 520 m and ∠D = 50° . How far is E
from D (accurate to 0.1 m) so that A, C, E lie in the same line?
A
B
C
E
140°
9.8 m
B
i = 1：1.6
50°
D
A
(No. 3)
5.8 m
α
D
(No. 4)
- 277 -
C
Practice
4. As in diagram, the cross section of the foundation of a railway is
an equilateral trapezium ABCD, From the figures provided in the
diagram calculate the width of base AD (accurate to 0.1m) and
the angle of slope α .
Exercise 10
1. Find the four trigonometric values of ∠A and∠B for each of the
following triangles.
B
C
B
8
C
55
73
16
10
(1)
b
?
c
b
What trigonometric function of ∠B has the value ?
c
a
(3) What trigonometric function of ∠A has the value ?
b
a
What trigonometric function of ∠B has the value ?
b
(2) What trigonometric function of ∠A has the value
A
A
(2)
(No. 1)
C
73
B
A
(3)
2. In the right-angled triangle ABC ( C = 90° ):
(1) Given a = 9 , c = 15 c = 15 , find the four trigonometric
values of ∠A.
(2) Given b = 21 , c = 29 , find the four trigonometric values of
∠A.
(3) Given a = 2 , b = 6 , find the four trigonometric values of
∠A and ∠B.
3. Determine whether a triangle with sides 8 cm, 15 cm, 17 cm is a
right-angled triangle. If the answer is yes, find the four
trigonometric values of the angle opposite to the shortest side.
4. In the right-angled triangle ABC ( C = 90° ):
a
(1) What trigonometric function of ∠A has the value ?
c
a
What trigonometric function of ∠B has the value ?
c
- 278 -
5. Solve the right-angled triangles given the following information
(Without using a calculator).
(1) c = 10 , A = 45° ;
(2) a = 6 , B = 30° ;
(3) a = 50 , c = 50 2 ;
(4) a = 8 5 , b = 8 15 .
6. Solve the right-angled triangles using the calculator, given the
following information:
(1) c = 8.035 , A = 38°19′ ;
(2) b = 7.234 , A = 7°20′ ;
(3) a = 25.64 , b = 32.48 .
7. Given that the apex angle of an isosceles triangle is 78°4′ , and
its height is 28.5 cm, find the length of the two sides (keep 3
significant figures for the answer).
8. As in the diagram, at a point A, a distance of 150 m from a tower,
a clinometer measures the angle of elevation of B to be 30°12′ . It
is known that the clinometer has height AD = 1.5 m. Find the
height of the tower (accurate to 0.1 m).
B
140
30°12′
α
D
E
83
A
150
124
150 m
(No. 8)
(No. 9)
- 279 -
9. In the diagram, the gradient angle α has to be calculated. Find
α according to the measurements shown.
10. As in the diagram, the cross-section of a dam is a rhombus
ABCD，Calculate (i) the angles of slopes α and β according to the
figures in the diagram, (ii) the width of the bottom AD and (iii)
the length of slope AB (accurate to 0.1 m).
B 2.8 m C
i =1：2.5
A
7.5 m
4.2 m
β
α
E
(No. 10)
D
11. As in the cross-sectional diagram, a mould is constructed from a
bar by excavating the material to a depth of 0.6 m from the centre
of the bar to form a basin (refer to the portion marked as (III) in
the diagram), and piling the excavated material on the two sides
(refer to the portions marked as (I) and (II) in the diagram) .
Given the slope of the basin is 1：1.5, the width of the bottom
of the basin BC is 0.5 m, find
(1) Area of the cross-section ABCD (isosceles trapezium).
(2) What volume (in cubic meters) of material is excavated from
the centre in building a mould 2 m in length?
(I)
D
A
i =1：1.5
B
C
(III)
(No. 11)
(II)
III. Solving general triangles
15.7
Trigonometric functions of obtuse angles
In production work and scientific research, there often arise
problems involving triangles which are not right-angled triangles. To
study the relation between sides and angles of a general triangle, let
us study the trigonometric functions of angle α when
90° ≤ α < 180° .
When α = 90° ,the terminal side of ∠α
y
coincides with the positive half y-axis Oy
(diagram 15-17). In this case for any point P (x, y)
P (x, y)
on the terminal side of ∠α , x = 0 , y = r = OP .
r
Therefore
90°
y
x
x
sin 90° = = 1 ,
cos 90° = = 0
O
r
r
Diagram 15-17
x
tan 90° = undefined , cot 90° = = 0
y
When 90° < α < 180° , ∠α has terminal
y
side in the second quadrant (diagram 15-18).
For any point P (x, y) on the terminal side of
P (x, y)
∠α , we have x < 0 , y > 0 , r = OP > 0 .
y
r
Therefore
α
y
x
x
sin α = > 0 ， cos α = < 0
M x O
r
r
Diagram 15-18
y
x
tan α = < 0 ， cot α = < 0
x
y
We know that the trigonometric values of acute angles are
positive (why?), but for obtuse angles, all trigonometric values are
negative except for the sine value.
【Example 1】Given that the terminal side of ∠α passes through
the point P( −3 , 4), find the four trigonometric values
of ∠α . (Diagram 15-19)
- 280 -
- 281 -
y
Solution ∵
∴
∴
x = −3 , y = 4
r = (−3) 2 + 4 2 = 25 = 5
y 4
x
3
sin α = = , cos α = = − ,
r 5
r
5
y
4
x
3
tan α = = − , cot α = = − .
y
x
3
4
P( −3 , 4)
α
O
Diagram 15-19
x
Given an obtuse angle, how can we find the trigonometric
values?
The trigonometric values of an acute angle can be looked up
from tables. If somehow we can transform the problem of finding
trigonometric values of an obtuse angle to that of an acute angle, then
the problem is solved.
It is easily seen that any obtuse angle can be expressed in the
form 180° − α where α is an acute angle. For example,
120° = 180° − 60° . Let use investigate the relation between the
trigonometric values of 180° − α and α .
As in diagram 15-20, pick any point
y
P(x, y) on the terminal side of obtuse
P (x, y)
angle 180° − α . Let OP = r . Choose a
P1 ( x1 , y1 )
r
r
point P1 ( x1 , y1 ) on the terminal side of
angle α so that OP1 = r . Since OP
180° − α
α
x
and OP1 make equal angles with the
O
Diagram 15-20
y-axis and OP = OP1 , P and P1 are
symmetric about the y-axis. The coordinates of the two points are
related. Thus
x = − x1 , y = y1
y y
∴ sin(180° − α ) = = 1 = sin α
r r
x −x
cos(180° − α ) = = 1 = − cos α
r
r
- 282 -
y
y
= 1 = − tan α
x − x1
x −x
cot(180° − α ) = = 1 = − cot α
y
y1
When α is acute, we have
sin(180° − α ) = sin α ,
cos(180° − α ) = − cos α
tan(180° − α ) =
tan(180° − α ) = − tan α , cot(180° − α ) = − cot α
It is recommended to memorize these formulas, because the
formulas have wide applications.
【Example 2】Find the values of (1) sin120° ; (2) cos158°14′ ;
(3) tan135° ; (4) cot150°18′ .
3
Solution (1) sin120° = sin(180° − 60°) = sin 60° =
;
2
(2) cos158°14′ = cos(180° − 21°46′)
= − cos 21°46′
= −0.9287
(3) tan135° = tan(180° − 45°) = − tan 45° = −1 ;
(4) cot150°18′ = cot(180° − 29°42′)
= − cot 29°42′
= −1.753
5
, 0° < α < 180° , find α .
6
(2) Given cos α = −0.8728 , 0° < α < 180° , find α .
5
Solution (1) Given sin α = ≈ 0.8333 , 0° < α < 180° . Therefore
6
α can be acute, and it can also be obtuse. Using a
scientific calculator we get sin 56°27′ = 0.8333 .
∴ α1 = 56°27′
Also ∵ sin(180° − 56°27′) = sin 56°27′ = 0.8333
∴ α 2 = 180° − 56°27′ = 123°33′
【Example 3】(1) Given sin α =
- 283 -
There are two solutions,
α1 = 56°27′ , α 2 = 123°33′ .
(2) It is known that cos α is negative, 0° < α < 180° ,
therefore α is obtuse.
Let α = 180° − θ , where θ is acute. Then
cos α = cos(180° − θ ) = − cos θ = −0.8728 .
So cos θ = 0.8728 . By using a scientific calculator,
we get θ = 29°13′ , therefore
α = 180° − 29°13′ = 150°47′ .
Practice
1. Given that the terminal side of α passes through the points for
the four cases below, find the four trigonometric values of α .
for each case.
(1) ( −2 , 2); (2) ( −1 , 3 );
(3) ( −2 , 5 ); (4) (0, 3).
2. Evaluate
(1) sin135° , cos120° , tan150° , cot150° ;
(2) sin118°8′ , cos100°24′ , tan 95°12′ , cot151°42′ ;
(3) cos123°26′ , sin 90°10′ , cot134°43′ , tan172°21′ .
3. Given 0° < θ < 180° , find the value of θ .
1
3
(1) sin θ = ;
(2) sin θ = 0.6517 ; (3) cos θ = −
;
2
2
3
(4) cos θ = −0.3541 ; (5) tan θ = −3.566 ; (6) cot θ = −
.
3
15.8
be (x, y). From the definitions of
y
trigonometric functions, no matter
B (x, y)
what type of angle A is, we always
x
y
a
have
= cos A ,
= sin A , and y
c
c
c
so x = c cos A , y = c sin A i.e.
x O A
C (b, 0)
b
the coordinates of B is ( c cos A ,
Diagram 15-21
c sin A ). Also the coordinates of C
is (b, 0), from the distance between two points formula,
x
a = BC = (b − c cos A) 2 + (−c sin A) 2
Squaring both sides,
a 2 = (b − c cos A) 2 + (−c sin A)2
Simplifying the right hand side,
(b − c cos A)2 + (−c sin A) 2 = b 2 − 2bc cos A + c 2 cos 2 A + c 2 sin 2 A
= b 2 + c 2 (sin 2 A + cos 2 A) − 2bc cos A
= b 2 + c 2 − 2bc cos A
Therefore
a 2 = b 2 + c 2 − 2bc cos A
(1)
If vertex B of △ABC is used as the origin to build a rectangular
coordinate system as in diagram 15-22, we can similarly prove
b 2 = c 2 + a 2 − 2ca cos B
(2)
2
2
2
c = a + b − 2ab cos C
(3)
y
y
C (x, y)
The Cosine Rule
A (x, y)
c
a
In this and the next section we shall prove two important
theorems – the Cosine Rule and the Sine Rule, and show how they
can be used to solve general triangles.
Use vertex A of triangle ABC as the origin, ray AC as the
positive x-axis to construct a coordinate system, as in diagram 15-21.
Now vertex B can be viewed as a point on the terminal side of angle
A, its distance from the origin being r = c . Let the coordinates of B
- 284 -
b
OB
c
A (c, 0)
b
x
O C
a
x
B (a, 0)
Diagram 15-22
From this, we obtain an important theorem relating the three
sides and an angle of an arbitrary triangle.
- 285 -
a 2 + b 2 − c 2 72 + 102 − 62 113
=
=
≈ 0.8071 .
2ab
2 × 7 ×10
140
Using calculator, we get C ≈ 36° .
(3) B = 180° − ( A + C ) ≈ 180° − (44° + 36°) = 100° .
Cosine Rule:The square of any side of a triangle is equal to the
sum of squares of the other two sides; minus twice
the product of their sides and the cosine of their
included angle.
a 2 = b 2 + c 2 − 2bc cos A
b 2 = c 2 + a 2 − 2ca cos B
c 2 = a 2 + b 2 − 2ab cos C
If one of the angles of triangle ABC is a right angle, for example,
C = 90° , then cos C = 0 . From the Cosine Rule we get c 2 = a 2 + b 2 ,
which is the Pythagoras Theorem. From this we see that the Cosine
Rule is an extension of the Pythagoras Theorem, and the Pythagoras
Theorem is a special case of the Cosine Rule.
From (1), (2), (3),
b2 + c 2 − a 2
cos A =
2bc
2
c + a 2 − b2
cos B =
2ca
2
a + b2 − c 2
cos C =
2ab
The Cosine Rule can be used to solve the following two
problems:
(1) Finding the three angles given three sides.
(2) Given two sides and their included angle, finding the third
side and the other two angles.
【Example】 In △ABC, given that a = 7 , b = 10 , c = 6 , find A,
B and C (accurate to 1° ).
b 2 + c 2 − a 2 102 + 62 − 7 2 87
Solution (1) cos A =
=
=
= 0.7250 .
2bc
2 × 10 × 6
120
Using calculator, we get A ≈ 44° .
- 286 -
(2) cos C =
Note: In the example we use the Cosine Rule to find first the two
acute angles, and then use the angle sum of triangle to find B,
B = 100° is an obtuse angle. If angle B is found using the
Cosine Rule, since B is obtuse, cos B will be negative. In
general, to find the angles of a triangle from the three sides,
we can follow the rule that “the smaller side has a smaller
opposite angle” to calculate first the angle opposite to the
smallest side. The angle thus found must be acute. (Why?).
Practice
1. In triangle ABC:
(1) Given b = 8 , c = 3 , A = 60° , find a;
(2) Given a = 20 , b = 29 , c = 21 , find B;
(3) Given a = 3 3 , c = 2 , B = 150° , find b;
(4) Given a = 2 , b = 2 , c = 3 + 1 , find A.
2. Solve the triangle given the following information.:
(1) a = 31 , b = 42 , c = 27 ;
(2) a = 9 , b = 10 , c = 15 .
15.9
The Sine Rule
As in diagram 15-23, align the
vertex A of △ABC with the origin
and align the side AC with the
positive x-axis to form a rectangular
coordinate system. We know from
the previous section that the
coordinates of B is ( c cos A , c sin A ).
It is easy to see that the height
- 287 -
y
B ( c cos A, c sin A )
a
c
E
O A
b
Diagram 15-23
C
x
corresponding to side BC is BE, and is equal to c sin A . Therefore the
area of △ABC is
1
1
S△ = × AC × BE = bc sin A 。
2
2
Similarly (see diagram 15-22):
1
1
S△ = ca sin B , S△ = ab sin C .
2
2
From this, we get the formula for the area of a general triangle:
1
1
1
bc sin A = ca sin B = ab sin C
2
2
2
That is to say, the area of a triangle is equal to half of the
product of any two sides times the sine of their included angle.
From the equations
1
1
1
bc sin A = ca sin B = ab sin C
2
2
2
1
divide by abc , we get
2
sin A sin B sin C
=
=
。
a
b
c
Hence we get for any triangle an important theorem on the
relations between the sides and angles:
S△ =
Sine Rule:
In a triangle, the ratios of the sides to the sine
values of their opposite angles are equal.
a
b
c
=
=
sin A sin B sin C
If one of the angles of △ABC is a right angle, for example,
a
C = 90° , then sin C = 1 . From the Sine Rule we get sin A = ,
c
b
sin B = . These are exactly the relations between sides and angles
c
of a right-angled triangle in Section 15.4.
- 288 -
By using the Sine Rule and the Sum of Angles Theorem, two
kinds of problems for a general triangle can be solved:
(1) Given two angles and a side, find the other two sides and
angle.
(2) Given two sides and the opposite angle of one of the sides,
find the opposite angle of the other side (Hence finding the
third side and its opposite angle).
【Example 1】In △ABC, given that c = 10 , A = 45° , C = 30° ,
find b amd S△ (correct to two significant figures).
b
c
Solution (1) ∵
=
, B = 180° − ( A + C )
sin B sin C
c sin[180° − ( A + C )]
∴ b=
sin C
10sin[180° − (45° + 30°)]
=
sin 30°
10sin 75°
=
1
2
≈ 19.3
1
(2) S△ = bc sin A
2
1
≈ × 19.3 × 10 × sin 45° ≈ 68
2
【Example 2】In △ABC, given that a = 20 , b = 28 , A = 40° ,
find B (accurate to 1° ) and c (correct to two
significant figures).
b sin A 28sin 40°
Solution (1) sin B =
=
≈ 0.8999 .
a
20
By using a calculator, B1 = 64° .
Also sin(180° − B1 ) = sin B1 , the obtuse angle
B2 = 180° − B1 also satisfies the conditions of the
problem, we get B2 = 180° − 64° = 116° .
- 289 -
Solution (1) Given b < a , therefore B < A , ∠B is also an acute
angle.
b sin A 50sin 38°
sin B =
=
≈ 0.5131 , by using a
a
60
calculator, B = 31° .
a sin C a sin[180° − ( A + B) a sin( A + B )
=
=
(2) c1 =
sin A
sin A
sin A
60 i sin(38° + 31°) 60 i sin 69°
=
=
≈ 91
sin 38°
sin 38°
(2) When B1 = 64° ,
a sin[180° − ( A + B1 )
c1 =
sin A
20 i sin[180° − (40° + 64°)]
=
sin 40°
20 i sin 76°
=
sin 40°
≈ 30
When B2 = 116° ,
a sin[180° − ( A + B1 )
c2 =
sin A
20 i sin[180° − (40° + 116°)]
=
sin 40°
20 i sin 24°
=
sin 40°
≈ 13
Think for a while: Can Cosine Rule be used to solve example 2?
In diagram 15-24, we first
draw ∠PAQ = 40° , then on AQ
let AC = b = 28 mm. Use C as
center, a = 20 mm as radius to
draw an arc. We see that the arc
meets AP at two points B1 , B2 .
Thus two triangles can be formed.
Both △ AB1C and △ AB2C
satisfy the conditions. This shows
example 2 has two solutions.
Q
C
C
a
A
a
c2 B2
C
a
b
b
b
A
B1
A
a < b sin A
No Solution
C
P
c1
Diagram 15-24
【Example 3】In △ABC, given a = 60 , b = 50 , A = 38° , find B
(accurate to 1° ) and c (correct to two significant
figures).
- 290 -
【Example 4】In △ABC, given a = 18 , b = 20 , A = 150° .
Solve this triangle.
Solution Here A is obtuse, a should be the longest side, but b > a ,
therefore there is no solution for the problem.
For examples 3 and 4, follow the method in diagram 15-24 and
try to draw triangles. What is the result?
In general, solving a triangle given two sides and an angle
opposite to one of the sides，there are 3 possible outcomes: (i) no
solution, (ii) 1 solution and (iii) 2 solutions. Diagram 15-25(a), (b)
illustrate that in solving △ABC given a, b and A, there are the
following cases:
(1) A is acute:
a
B
a = b sin A
One Solution
C
b
b
a a
A
a
A
B2
B1
b sin A < a < b
a≥b
Two Solutions
One Solution
Diagram 15-24 (a)
- 291 -
B
(2) A is a right or obtuse angle:
C
a
C
a
b
C
b
A
B
A
a≤b
No Solution
a>b
One Solution
60°
A
Diagram 15-24 (b)
B
6°20′
Diagram 15-26
Practice
1. In triangle ABC:
(1) Given c = 3 , A = 45° , B = 60° , find b, S△ .
(2) Given b = 12 , A = 30° , B = 120° , find a.
(correct to two significant figures)
2. Solve the triangle given the conditions (If there are solutions,
correct angles to 1° and correct sides to two significant
figures):
(1) a = 15 , b = 10 , A = 60° ;
(2) b = 40 , c = 20 , C = 25° ;
(3) b = 11 , a = 25 , B = 30° .
Analysis: The problem mathematically is this: In △ABC, given
AB = 1.95 m, AC = 1.40 m, ∠BAC = 60° + 6°20′ = 66°20′ ,
find the length of BC. Since two sides and an included
angle of △ABC is known, BC can be calculated using the
Cosine Rule.
Solution By the Cosine Rule,
BC 2 = AB 2 + AC 2 − 2 AB i AC cos A
= 1.952 + 1.402 − 2 × 1.95 × 1.40 cos 66°20′
Using a calculator, we get BC ≈ 1.89 m。
Answer: Shaft BC is 1.89 m in length.
15.10
【Example 2】To build a bridge over a river, before work
commences two foundations A, B are laid on the
opposite sides of the river (diagram 15-27). In order
to measure accurately the distance between A and B，
the surveyors set a base line BC on one bank, and
obtain the measurements BC = 78.35 m,
∠B = 69°43′ , ∠C = 41°12′ . Calculate the length of
AB. (accurate to 0.01m)
Applications of Triagonometric functions
Some examples of applications of solving triangles to real life
problems are given below.
【Example 1】The compartment of an automatic unloading truck
uses a fluid compression mechanism. In its design
the length of the shaft of the oil pump BC has to be
calculated. It is given that the greatest elevation
angle of the compartment is 60° , the distance
between the top of the pump and the fulcrum of the
compartment is 1.95 m; AB makes an angle of 6°20′
with the horizontal; length of AC is 1.4 m. Calculate
the length of BC (keep three significant figures for
the answer).
- 292 -
A
B
Diagram 15-27
- 293 -
C
Solution In △ ABC, it is given that a = 78.35 , B = 69°43′ ,
C = 41°12′ .
A = 180° − ( B + C ) = 180° − (69°43′ + 41°12′) = 69°5′
From the Sine Rule, we have
a sin C 78.35 × sin 41°12′
c=
=
.
sin A
sin 69°5′
By using a calculator, we get c ≈ 55.26 m.
Answer: Foundations A,B are 55.26 m in distance.
【Example 3】Diagram 15-28 is an illustration of a connecting rod
mechanism. When crank CB rotates around C, acting
through the connecting rod AB, the piston moves
back and forth in a straight line. When the crank is at
CB0 , it is in a straight line with the connecting rod
and the endpoint of the rod is at A0 . Suppose that
the connecting rod AB is 340mm long and the crank
CB is 85 mm long, find the distance the piston
moves when the crank CB0 rotates 80° in the
clockwise direction. (i.e. distance A0 A moved by
the endpoint A of the connecting rod) (accurate to
1mm)
B
B
80°
A0
B0
A
C
A0
(1)
B0
A
80°
C
Solution In △ABC, from the Sine Rule,
BC sin C 85 × sin 80°
sin A =
=
≈ 0.2462 .
AB
340
By using a calculator, A ≈ 14°15′ .
From the Sine Rule,
AB sin B 340 × sin 85°45′
AC =
=
.
sin C
sin 80°
By using a calculator AC ≈ 344.3 mm.
Therefore
A0 A = A0C − AC
= ( AB + BC ) − AC
= (340 + 85) − 344.3
= 80.7
≈ 81 mm
Answer: When the crank rotates clockwisely 80° from CB0 ,
the distance moved by the piston is 81 mm.
Practice
1. Refer to the diagram, to measure the distance between two
points A, B (which are obscured from each other, and not
accessible directly), a suitable point C is chosen on the ground;
with the measurements AC = 213.4 m, BC = 252.1 m,
∠ACB = 50°13′ . Calculate the length of AB.
(2)
Diagram 15-26
Analysis: Since A0 A = A0C − AC , and given A0C = AB + BC = 340
+ 85 = 425 mm, the problem is solved once we find the
length of AC. In △ABC, two sides and the opposite angle
of one of the sides are known, AC can be found by the
Sine Rule.
- 294 -
A
B
C
(No. 1)
- 295 -
Practice
Practice
2. As in diagram, three holes A, B, C are to be drilled on a certain
part of a device. It is known that the distances between centres
are : AB = 112.5 mm, BC = 75.4 mm, AC = 114.2 mm. If holes
A, B are completed and the driller is at hole B，how far along BA
and how far perpendicular to BA the driller has to move in order
to drill hole C ( i.e. find x and y in the diagram，accurate to 0.1
mm)?
4. The diagram is an illustration of a connecting rod mechanism.
When the crank is horizontal at OB，the end of the connecting
rod is at Q. When the crank rotates from OB to OA in a
clockwise direction an angle α , the endpoint of the connecting
rod is at P. The distance between P and Q is x. Given OA = 25
cm, AP = 125 cm，find the value of x for each of the following:
(1) α = 50° ；
(2) α = 90° ；
(3) α = 135° ；
(4) OA ⊥ AP ；
C
A
x
α
y
B
x
Q
A
3. As in diagram, to measure the height of a chimney AB, from two
points C, D on the same horizontal level as the bottom of the
chimney, the angles of elevation of the chimney are α = 35°12′
and β = 49°28′ respectively. The distance between C, D is
11.12 m. Given that the height of the clinometer is 1.52 m, find
the height of the chimney.
B
C
α
D1
β
A1
D
A
(No. 3)
- 296 -
O
(No. 4)
(No. 2)
C1
B
P
C
Exercise 11
1. Find the trigonometric values:
(1) sin158°54′ , cos172°36′ , tan105°6′ , cot 91°42′ ;
(2) sin136°27′ , cos118°38′ , tan155°56′ , cot142°19′ .
2. Write the expressions in the form of trigonometric functions of
acute angles α or β .
(1) sin (180° − α ) ;
(2) cos (180° − β ) ;
(3) tan (90° − α ) ;
(4) cot (180° − β ) .
3. Suppose A, B, C are the internal angles of a triangle, prove that:
(1) sin ( A + B ) = sin C ;
(2) cos ( B + C ) = − cos A .
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4. Find the internal angle A of a triangle which satisfies the
equations.
2
1
(1) sin A =
, cos A = − , tan A = −1 , cot A = − 3 ;
2
2
(2) sin A = 0.2529 , cos A = −0.9756 , tan A = −1.998 .
5. In △ABC:
(1) Given a = 49 , b = 26 , C = 107° , find c and B.
(2) Given a = 84 , b = 56 , c = 74 , find A.
(3) Given c = 68 , A = 34° , B = 56° , find a and S △ .
(Correct angles to 1° , correct sides and areas to 2 significant
figures)
6. Solve the triangles according given the following information.
(Correct angles to 1° , correct sides to 2 significant figures)
(1) b = 26 , c = 15 , C = 23° ;
(2) b = 54 , c = 39 , C = 115° .
7. From point O in a quadrilateral
D
playground ABCD , the following
measurements are obtained:
A
OA = 39.8 m, OB = 36.7 m,
102°30′
OC = 42.3 m, OD = 44.1 m,
88°
∠AOB = 88° , ∠BOC = 76° ,
76°
C
∠COD = 102°30′ , find the area of the
B
(No.
7)
playground (accurate to 2 significant
figures).
8. Given that the adjacent sides of a parallelogram are 4 6 cm
and 4 3 cm in length and their included angle is 45° ，find the
lengths of the diagonals and area of the parallelogram.
9. In the diagram, parallelogram ABCD has diagonal AC = 57 cm.
The angles AC making with adjacent sides AB and AD are
respectively α = 27° and β = 35°，find AB and AD (accurate to 1
cm).
- 298 -
D
A
C
D
β
α
A
E
B
C
B
(No. 9)
(No. 10)
10. To dig a tunnel, the distance between its two ends D,E is to be
measured. A suitable point C on one side of the mountain is
chosen for this purpose. Measurements are obtained: CA = 482.8
m, CB = 631.5 m, ∠ACB = 56°18′ . Also the distances of A, B
to the entry points of the tunnel are AD = 80.12 m, BE = 40.24 m
(A, D, E, B are on the same straight line). Calculate the length of
tunnel DE.
11. As in diagram, a ship sails due East with speed 32.2km/hr. At a
point A a beacon tower S is sighted in the direction of E 20° S
from the ship. Half an hour later it reaches B, and the tower is
sighted in the direction of E 65° S. Find the distance of the tower
from B (accurate to 0.1km).
B
20°
A
B
α
C
β
65°
N
W
S
E
A
S
D
(No. 11)
(No. 12)
12. As in diagram, from the the top of a tower B on a mountain top，A
point A on the ground has an angle of depression α = 54°40′ ，
from point C at the bottom of tower the angle of depression of A
is β = 50°1′ . Given that the height of the tower BC is 27.3 m, find
the height of the mountain CD (accurate to 1 m).
- 299 -
13.
As in diagram, a boat sails with a
speed of 35km/hr. on a bearing of
148° (it is the direction measured
by the counterclockwise angle made
with the North). To locate its
position, the boat keeps measuring
the bearing of a watch tower A. At B,
the bearing of the watch tower A is
126° . After half an hour of sailing,
the boat reaches point C and the
bearing of A becomes 78° . Find
the distance of A from the boat
when it reaches point C (accurate to
1km).
N
There are the following identities between the trigonometric
functions of an angle, the trigonometric functions of its
complimentary angle and the trigonometric functions of its
supplementary angle:
148°
B
126°
(1) sin(90° − α ) = cos α , cos(90° − α ) = sin α
tan(90° − α ) = cot α , cot(90° − α ) = tan α
N
78°
(2) sin(180° − α ) = sin α , cos(180° − α ) = − cos α
A
C
(No. 13)
tan(180° − α ) = − tan α , cot(180° − α ) = − cot α
III. The trigonometric values of some special angles can be
found from their corresponding right-angled triangles, as follows:
30°
45°
60°
90°
Chapter Summary
Sine
1
2
2
2
3
2
1
I. This chapter mainly teaches the fundamental concepts of
trigonometric functions and the application of trigonometric
functions to solve problems on triangles.
Cosine
3
2
2
2
1
2
0
II. Let angle α has vertex at the origin, its initial side
coinciding with the positive x-axis. P(x, y) is any point on the
terminal side with OP = r . The trigonometric functions of angle
α are
y
The Sine of α is sin α =
r
x
The Cosine of α is cos α = ,
r
y
The Tangent of α is tan α =
x
x
The Cotangent of α is cot α = .
y
Tangent
3
3
1
- 300 -
3
Undefined
3
0
3
The trigonometric values of non-special acute angles can be
found using scientific calculator or by looking up trigonometric
tables. The trigonometric values of obtuse angles can be found by
using the relations between supplementary angles 180° − α and α ,
whereby the problem is transformed into one that requires only the
finding of the trigonometric values of the acute angle α .
Cotangent
3
1
Conversely, if the trigonometric value of an acute or obtuse
angle is known, we can find the value of the angle.
- 301 -
IV. Let α be an acute angle of a right-angled triangle. Then
Opposite side of α
Adjacent side of α
sin α =
, cos α =
Hypotenuse
Hypotenuse
Opposite side of α
Adjacent side of α
, cot α =
Adjacent side of α
Opposite side of α
By using these equations and the complementary relationship
between the two acute angles of a right-angled triangle, any
right-angled triangle can be solved. The key is to choose the
appropriate equation that contains no more than one unknown in its
equation.
V. For a general triangle ABC, the sides and angles have the
following relations:
(1) The Law of Cosines or the Cosine Rule
 a 2 = b 2 + c 2 − 2bc cos A
 2
2
2
 b = c + a − 2ca cos B
 c 2 = a 2 + b 2 − 2ab cos C

tan α =
(2) The Law of Sines or the Sine Rule
a
b
c
=
=
sin A sin B sin C
To solve a general triangle, there are 4 cases:
(1) 3 sides are known.
(2) 2 sides and their included angle are known.
(3) Two angles and one side are known.
(4) Two sides and an opposite angle of one of the sides are
known.
In general, cases (1), (2) can be solved by the Cosine Rule and
the angle sum formula. Cases (3), (4) can be solved by the Sine Rule
and the angle sum formula. For cases (1),(2),(3), there can be two
different outcomes, namely (i) one solution or (ii) no solution at all.
For case (4), there can be three different outcomes, namely (i) two
solutions, (ii) one solution or (iii) no solution at all. In a triangle,
when given 3 elements whether there are known sides or known
angles (note: at least one of the known elements must be a known
- 302 -
side), we can find out all the unknown sides and angles of the
triangle.
By being able to quantitatively evaluate the relationship of all
the elements of a triangle, we have gained greater understanding of
triangles.
Revision Exercise 15
1. Given the terminal side of angle α passes through the following
points, find the four trigonometric values of angle α .
(1) ( −8 , 6);
(2) (5, 7 );
2
(3) ( − 2 , 3 );
(4) (0, ).
3
2. In the right-angled triangle ABC, angle C is a right angle, CD is
the height corresponding to the hypotenuse AB. Write down for
angle A the ratios of sine, cosine and tangent. How many ratios
are there?
3. Evaluate:
(1) tan 2 150° + 2 sin 60° + tan 45° sin 90° − tan 60° + cos 2 30° ;
3
(2) cos 60° − sin 2 135° + tan 2 30° + cos 2 30° − sin 30° ;
4
sin 90°
(3)
;
tan 45° − sin120°
tan 30° cos135° sin 60°
.
(4)
cot120° sin150°
4. Evaluate the following trigonometric values:
(1) sin162°21′ , cos103°16′ , tan 93°53′ , cot145°3′ ;
(2) sin109°43′ , cos167°4′ , tan151°32′ , cot100°32′ .
5. Find the interior angles α which satisfy the equations
( 0° < α < 180° ):
3
(1) sin α =
, cos α = −0.6581 , tan α = 35.43 ;
2
(2) sin α = sin15° , cos α = − cos 45° , − cot α = cot115° .
- 303 -
6. Given that A,B,C are the interior angles of a triangle, it is requied
to prove the following:
A+ B
C
B+C
A
(1) sin
= cos ;
(2) tan
= cot .
2
2
2
2
7. A dam is to be widened 2.0m on its opposite side to the water.
The gradient changes from the origin 1: 2 to 1 : 2.5. The original
slope has length BD=13.4m. The dam is 90m long. Find the
volume of earth needed to complete the project (correct to two
significant figures).
C
1：2.5
A
2.0
D
1：2
B
E
F
(No. 7)
8. In the right angle triangle ABC, angle A = 32°20′ , AT is the angle
bisector of angle A with AT = 14.7 cm. Find the lengths of the
sides BC and the hypotenuse AB (correct to 3 significant figures).
9. The wing of a certain model of plane is as in the diagram.
According to the figures provided, calculate the lengths of AC,
BD and AB. (correct to 3 significant figures)
11. Trapezium ABCD has the top DC = 15.3 cm, the bottom AB =
24.2 cm, one side AD = 12.0 cm, another side BC = 11.7 cm. Find
the angles of the trapezium and its area (correct to 3 significant
figures).
12. Given that △ABC has area 12 cm2, sides a = 6 cm, b = 8 cm，find
their included angle C. Draw a diagram to show the solution.
13. In △ABC, AD is the angle bisector of A. Use the Sine Rule to
prove that
BD AB
=
DC AC
14. As in diagram, between A, B there is a hill and a river. To find the
distance of AB, a point D is picked so that AD can be measured
directly and B, D are in sight of each other. Furthermore a point C
on AD is picked so that B, C are in sight of each other. It is
measured that AC = 149.46 m, CD = 52.61 m, ∠ADB = 108°15′ ,
∠ACB = 120°25′ . Find AB.
B
N
N
A
3.4 m
D
D
15.8
80°
30°
A
105°
？
45°
C
45°
C
C
(No. 14)
A
A
B
(No. 15)
C
B
10. The take-up lever of a sewing machine has shape as in the
diagram. In finishing its production, the distance between the two
holes A,C has to be calculated. Given that BC = 60.5 mm, AB =
15.8 mm, ∠ABC = 80° , find the length of AC (correct to 3
significant figures).
15. A fishing boat is in distress at sea and sends out SOS signals. A
battleship at A detects that the fishing boat is at C in the bearing
of 45° and at a distance of 10km away, sailing at a speed of
9km/hr. in a bearing of 105° , towards a small island B. The
battleship immediately goes to the rescue and sails at a speed of
21km/hr. Find the sailing direction of the battleship and the time
it takes to reach the fishing boat. (Hint: Let the battleship reached
B in x hours after receiving the signal).
- 304 -
- 305 -
5m
60.5
B
(No. 9)
(No. 10)
16. Use the Cosine Rule to prove: The sum of squares of the
diagonals of a parallelogram is equal to the sum of squares of the
four sides.
17. Using the area formula of a triangle ( Heron’s Formula)
S△ = s ( s − a)( s − b)( s − c)
1
( s = (a + b + c) , a, b, c are the lengths of the three sides),
2
calculate the area of the triangle S △ given the following
information:
(1) a = 20 , b = 13 , c = 21 ;
(2) a = 17 , b = 21 , c = 10 .
This chapter is translated to English by courtesy of Mr. Robert
WONG, and reviewed by courtesy of Mr. SIN Wing Sang, Edward.)
- 306 -