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Mathematics A30
Module 2
Lesson 19
Mathematics A30
Angles and Trigonometric Ratios
Part II
515
Lesson 19
Mathematics A30
516
Lesson 19
Angles and Trigonometric Ratios
Introduction
Lesson 19 concludes the section on angles for Mathematics A30. Using a calculator,
problems will be easier to solve compared to using tables. Calculators show one solution
for solving an equation for angles when given the trigonometric ratio. Using the cast rule
and reference angles, that you learned last lesson, other solutions will be found.
Problem solving with trigonometric ratios will once again be discussed. Some problems
now involve using two right triangles to obtain the solution.
Two special right triangles will be introduced. The special qualities of the two triangles
are that exact values for the trigonometric ratios can be obtained. Many professions, such
as surveying, construction, astronomy, rely on trigonometric ratios.
Mathematics A30
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Lesson 19
Mathematics A30
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Lesson 19
Objectives
After completing this lesson, you will be able to
•
determine the values for the trigonometric ratios by using a calculator.
•
apply the trigonometric ratios to problems involving right triangles.
•
determine the relationships among the sides of each special right triangle
(45° – 45° – 90° and 30° – 60° – 90° ).
•
calculate the length of the missing sides of the special right triangles when given
the exact value of one side.
Mathematics A30
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Lesson 19
Mathematics A30
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Lesson 19
19.1
Angles Associated with a Given Ratio
In lesson 18, you worked with a given angle, and using a calculator or the trigonometric
tables you were able to find any of the six trigonometric ratios.
Your calculator should be in degree mode for this lesson.
Find:
sin 55 =
cos 103 =
tan 217 =
csc 324 =
sec 117 =
cot 18 =
The inverse function keys on a scientific calculator can find the measure of an angle when
the trigonometric function is given.
The inverse function keys are:
•
•
•
1
sin
1
cos
1
tan
Find .
sin  = 0.8192
cos  =  0.2250
tan  = 0.7536
=
=
=
csc  =  1.7013
sec  =  2.2067
cot  = 3.0777
=
=
=
Mathematics A30
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Lesson 19
Method #1
Using a calculator
The keystroke pattern for finding  for sin  = 0.8192 is:
CLEAR
DISPLAY:
Method #2
2nd sin1 ( 0.8192 )
55.0048
ENTER
Using the Trigonometric Table
After searching through the sin column of the chart to locate 0.8192, follow across the
table to the left to find the value for  is 55 .
If you are using the trigonometric table remember to use reference angles to locate
angles over 90 .
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Lesson 19
Example 1
Determine all possible values between 0° and 360 for  in each of the
following.
a)
sin  = 0.6363
b)
cos  = 0.9200
c)
tan  =  1.3850
Solution:
a)
 S
A 
T
•
Sine is positive in the first and second
quadrant.
C
The angle set-up will look similar to:
y
x
•
sin  = 0.6363
 = 39.52°
Use the following keystroke pattern.
CLEAR
DISPLAY:
2nd sin1 ( 0.6363 )
39.5165
ENTER
 = 39.52 (to the nearest hundredth degree)
Mathematics A30
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Lesson 19
•
We now have an angle in the first quadrant ( = 39.52 ).
y

•
x
For the angle in the second quadrant, 39.52° becomes the reference angle.
y
1
4
0
.4
8

3
9
.5
2

x
The values that satisfy sin  = 0.6363 are 39.52 and 140.48 .
Check:
(with your calculator)
sin 39.52 = 0.6363 ()
sin 140.48 = 0.6363 ()
b)
cos  = 0.9200
•
Cosine is positive in quadrants
one and four.
S
A 
T
C 
cos  = 0.9200
 = 23.07°
Use the following keystroke pattern.
CLEAR
DISPLAY:
2nd cos1 ( 0.92 )
23.0739
ENTER
 = 23.07
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Lesson 19
•
We now have an angle in the first quadrant ( = 23.07 ).
y
23.07
•
x
For the angle in the fourth quadrant, 23.07 becomes the reference angle.
360  23.07 = 336.93
y
x
The values that satisfy cos  = 0.9200 are 23.07 and 336.93 .
Check:
(with your calculator)
cos 23.07 = 0.9200 ()
cos 336.93 = 0.9200 ()
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Lesson 19
c)
tan   1.3850
 S
T
A
Tangent is negative in quadrants two and four.
C 
tan θ =  1 .385
θ =  54 .17 
Use the following keystroke pattern.
CLEAR
DISPLAY:
2nd tan1 ((–)1.3850 )
 54.17
ENTER
 =  54.17
•
We now have an angle in the fourth quadrant  =  54.17 .
Reference Angles are always positive acute angles.
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Lesson 19
•
For the angle in the second quadrant,
 54 .17 = 54 .17  becomes the
reference angle.
180   54 .17  = 125 .83 
The values that satisfy tan  =  1.3850 are  54 .17  (or 305 .83  ) and 125 .83  .
Check:
tan  54 .17  =  1.3850 ()
tan 125 .83  =  1.3850 ()
The calculator will always display one angle and based on the cast rule and
reference angles, you will have to find the solution for the second angle.
In the last example you saw the solutions (angle) for the given trigonometric function
between 0 and 360 . The only difference for the next stage is that no restrictions will be
set.
The angle in standard position, coterminal with a given angle, with the smallest,
positive measure is called the principal angle.
1
is given, then one solution for  is the principal angle  = 30°
2
and another solution is the principal angle  = 180   30  = 150  .
If an equation like sin  =
Mathematics A30
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Lesson 19
1
2
 = 30
sin  =
•
The two principal angles satisfying the equation are called particular solutions to
the equation.
All other angles which are coterminal with 30 and 150 degree angles are also solutions
1
to the equation sin  = .
2
y
150°
x
510°
Check:
•
sin 390 = 0.5 ()
sin 510 = 0.5 ()
The general solution to the equation is the infinite set of all angles coterminal with
the particular solutions. The general solution may be written in a compact way.
General solutions to the equation sin  =
1
are:
2
 = 30 + n360, n 
Mathematics A30
 = 150 + n360, n 
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Lesson 19
Example 2
Find the particular solutions and the general solutions for csc  =
2
3
.
Solution:
csc  =
•
1
sin 
Particular Solution
1
2
=
sin 
3
1
2
=
sin 
3
sin  1  =   2  sin 
 sin    3 
 2 sin 
1=
3
 3
  2 sin   3 





  2 1  
3   2 



csc  =
Solve for sine.
3
2
sin   0.8660
sin  =
Find sin 1 .
 =  60
Use the following keystroke pattern.
CLEAR
DISPLAY:
2nd sin1 ((–)0.8660 )
 59.9971
ENTER
 =  60
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Lesson 19
Sine is negative in quadrant IV and quadrant III.
Quadrant III
Quadrant IV
 = 180 + 60
 = 300
 = 240
•
General Solution
 = 240 + n360, n  I
 = 300 + n360, n  I
Exercise 19.1
1.
If your calculator displays  0.9918 for sin 36 , is this correct? Why or why not? By
observation alone can you answer this question?
2.
Using your calculator, determine the value of  in each of the following.
a)
b)
c)
d)
e)
f)
sin 
cos 
tan 
sin 
cos 
tan 
Mathematics A30
= 0.7335
=  0.3882
= 3.247
=  0.4529
=  0.6525
= 1.1000
in Quadrant I
in Q. II
in Q. III
in Q. IV
in Q. III
in Q. I
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Lesson 19
3.
Determine all possible values between 0 and 360 for  in each of the following.
cos 
csc 
cot 
sec 
sin 
sin 
a)
b)
c)
d)
e)
f)
4.
= 0.7660
=  1 .9876
=  2 .3548
= 1.8182
=  0.3486
= 2.3500
1
sin 
1
sec  
cos 
1
cot  
tan 
csc  
Find the particular solution(s) and the general solution(s) for each of the following
equations.
3
2
csc A = 1
sin A  0 .6691
tan A = 0
sec A =  2
cos A 
a)
b)
c)
d)
e)
19.2
Problem Solving with Trig. Ratios
Tools you know and will need to solve the problems in this section.
•
Trigonometric Ratios
sin  =
opp
y
=
hyp
r
csc  =
hyp
r
=
opp
y
cos  =
adj
x
=
hyp
r
sec  =
hyp
r
=
adj
x
tan  =
opp
y
=
adj
x
cot  =
adj
x
=
opp
y
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r
y

x
Lesson 19
•
Pythagorean Theorem (for Right Angled Triangles)
2
2
leg + leg = hyp
2
2
2
x + y = r2
•
The sum of the angles in a triangle = 180
A
m A  m B  m C  180 
For a right angled triangle, the
measures of the two acute
angles add up to 90 . The acute
angles are complementary.
m A  m B  90   180 
m A  m B  90 
C
•
B
Math writing
•
Angles are written in CAPITAL letters.
measure of
m L  m M  90 
m N  90 
r 2  q 2  s2
•
Sides are indicated by the lower case of the letter of the opposite angle.
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Lesson 19
•
The angle of elevation is the angle between the horizontal line and the line of sight
looking up.
•
The angle of depression is the angle between the horizontal line and the line of
sight looking down.
Example 1
In the given right triangle, determine
the measures of the unknown side and
the measure of the unknown angles.
Solution:
Read the problem.
•
mG  90
g  17
f  15
•
Find mE, mF, e .
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Lesson 19
Develop a plan.
•
•
•
To find the measure of E , use the cosine ratio.
To find the measure of F , use the sine ratio.
To find side e, use the Pythagorean Theorem.
Carry out the plan.
adj
hyp
15
cos E =
17
cos E = 0.8824
Determine the measure of E .
cos E =
mE = 28.07
Use the following keystroke pattern.
CLEAR
DISPLAY:
15
÷
17
.8823529412
DISPLAY:
ENTER
2nd cos1 ( 0.8824 )
28.0668
ENTER
 = 28.07
opp
hyp
15
sin F =
17
sin F = 0.8824
Determine the measure of F .
sin F =
mF = 61.93
Use the following keystroke pattern.
CLEAR
DISPLAY:
15
÷
17
.8823529412
DISPLAY:
2nd sin1 ( 0.8824 )
61.9332
ENTER
ENTER
 = 61.93
Mathematics A30
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Lesson 19
Determine the length of side e.
leg
2
+ leg 2 = hyp
15 2
2
+ e 2 = 17 2
225 + e2 = 289
2
e = 64
e =  8 (Reject  8 )
e = 8
Use the following keystroke pattern.
CLEAR
DISPLAY:
17
(2nd
8
^
2
-
15
^
2 ) ENTER
x2)
Write a concluding statement.
The measure of side e is 8 units.
The measure of E is 28.07 .
The measure of F is 61.93 .
There are many different methods that could be used to obtain the same answers.
Check your method out!
Example 2
Solve ABC when given C = 22, B = 90, and a = 15.3 units.
Solution:
Solving a triangle refers to determining the values for all missing sides and angles.
Mathematics A30
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Lesson 19
Read the problem.
•
•
•
•
mC = 22
mB = 90
a = 15.3
Find mA, side c, side b .
Develop a plan.
•
•
•
•
Draw a diagram.
To find angle A, use the sum of all angles in a triangle must equal 180 .
To find side c, use the tangent ratio.
To find side b, use the cosine ratio.
Carry out the plan.
Draw a diagram.
Angle sum in triangles equals 180 degrees.
mA + mB + mC = 180
mA + 22 + 90 = 180
mA = 68
Determine the length of side c.
opp
adj
c
tan 22 =
15.3
c = tan 22  15.3 
tan 22 =
c = 0.4040  15.3 
c = 6.1816
Mathematics A30
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Lesson 19
Determine the length of side b.
adj
hyp
6.1816
cos 68 
b
cos A 
b cos 68 = 6.1816
Solve for b.
6.1816
cos 68
6.1816
b=
0.3746
b = 16.5016
b=
Check:
Use the Pythagorean Theorem.
a = 15.3
b = 16.5
leg 2 + leg 2
15 .3 2 + 6 .1816 2
= hyp 2
= 16 .5016 2
234 .09 + 38 .21 = 272 .3
c = 6.18
272 .3 = 272 .3  
Write a concluding statement.
m A = 68 
The missing values are:
b = 16 .05 units
c = 6 .18 units
Round off your answers to the second decimal place in the last step of your work.
This makes for a more accurate solution.
•
Some questions involve solving two right triangles before obtaining the necessary
measure.
Mathematics A30
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Lesson 19
Example 3
Determine the length of SP.
Solution:
Read the problem.
•
•
•
•
mRQS = 53
mRQP = 67
QR = 23 m
Find the length of SP using the trigonometric ratios in the two triangles
RQP and RQS .
Develop a plan.
•
•
•
Find the length of RP using RQP and the tangent ratio.
Find the length of RS using RQS and the tangent ratio.
Subtract the length of RS from the length of RP to find the length of SP.
Carry out the plan.
RP
23
RP = tan 67  23 
Find the length of RP.
tan 67  =
RP = 54 .1846 m
RS
23
RS = tan 53  23 
Find the length of RS.
tan 53  =
RS = 30 .5220 m
Mathematics A30
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Lesson 19
Find the length of SP.
SP = RP  RS
SP = 54.1846  30.5220
SP = 23.6626
SP = 23.66 m
Write a concluding statement.
The length of SP is 23.66 m.
Right angled triangles and their associated trigonometric ratios have many applications in
the real world. Let's try a few!
Example 4
Mr. Drummond is using a 4 m ladder to clean the windows on his house. His
ladder stands on level ground and leans against a wall at an angle of 52 .
A)
B)
How far is the foot of the ladder from the wall?
How high up the wall does the ladder go?
Solution:
Read the problem.
•
•
•
The ladder is 4 m long and this length represents the hypotenuse of the right
triangle.
The side adjacent to the 52 angle is the length from the foot of the ladder to the
wall.
The side opposite to the 52 angle is the length determining the height that the
ladder reaches up the wall.
Develop a plan.
•
•
Draw a diagram.
A) Use the cosine ratio to determine side a.
B) Use the sine ratio to determine side b.
Mathematics A30
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Lesson 19
Carry out the plan.
Draw a diagram.
A)
a
4
a = cos 52  4 
Write the ratio and solve for a.
cos 52 =
a = 2.4626
B)
b
4
b = sin 52  4 
Write the ratio and solve for b.
sin 52 =
b = 3.1520
Write a concluding statement.
The ladder is 2.5 m from the wall and reaches 3.2 m up the wall.
•
The method of solving right triangles is most useful for calculating distances which
cannot be measured directly.
Mathematics A30
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Lesson 19
Example 5
Using the information on the diagram,
determine the width of the river.
Solution:
Read the problem.
Find the width of the river x using the trigonometric ratios in the two triangles
ADC and BDC.
Develop a plan.
•
•
•
Find the length of DC using BDC and the tan ratio.
Find the length of AC using ADC and the tan ratio.
Subtract the length of BC from the length of AC to find the length of x.
Carry out the plan.
Find the length of DC.
22
DC
22
DC 
tan 33
tan 33 
22
0.6494
DC  33.8770
DC 
AC
DC
AC  (tan 49)(33.877)
Find the length of AC.
tan 49 
AC  (1.1504)(33.877)
AC  38.9710
x  38 .971  22
Find the length of x.
 16 .971
Write a concluding statement.
The width of the river is 16.97 m.
Mathematics A30
541
Lesson 19
Example 6
Dwight looks out his apartment window, directly across the street, at a
second building 39 m away. From Dwight's position, the angle of depression
to the bottom of the second building is 48 and the angle of elevation to the
top of the second building is 64 . What is the height of the building Dwight is
looking at?
Solution:
Read the problem.
•
•
•
•
The distance between the two buildings is 39 m.
The angle of depression is 48 .
The angle of elevation is 64 .
Find the height of the building.
Develop a plan.
•
•
•
•
Draw a diagram.
Use the tangent ratio to determine side a.
Use the tangent ratio to determine side b.
Add the two measures.
Carry out the plan.
Draw a diagram.
a
Buildin g
39 m
o
64
48 o
t ’s
gh n g
i
i
D w ild
u
B
b
Mathematics A30
542
Lesson 19
Use the tangent ratio to determine side a.
a
39
a = tan 64  39 
tan 64 =
a = 79.9618
Use the tangent ratio to determine side b.
b
39
b = tan 48  39 
tan 48 =
b = 43.3139
Add the two measures.
79.9618 + 43.3139 = 123.2757
Write a concluding statement.
The height of the building that Dwight is looking at is 123.28 m.
Exercise 19.2
1.
Solve each triangle.
a)
b)
2.
JKL when K = 90 , l = 5.3, and J = 32 .
Find the measure of x.
Mathematics A30
543
Lesson 19
3.
Jodi has a kite which is attached to a 100 m cord. When flying the kite one day, Jodi
notices that the kite has reached the end of this cord. Jodi is holding the cord 1.4 m
above the ground. Brayden, standing 72 m away from Jodi, states that the kite is
directly overhead. How high is the kite and what is the angle the cord makes with
the ground?
4.
A lawnmower has a handle 1.5 m long and is attached to the lawnmower at a point
20 cm above the ground. The manual states that the maximum efficiency in the
handling of the lawnmower is reached when the handle is at an angle of 60 with
respect to the ground. How high off the ground must the handle be held to achieve
this maximum efficiency?
5.
A forest ranger in a tower 200 m high sights two fires in the same line of sight with
angles of depression 10 and 15 . How far apart are the fires? How far away is the
nearer fire to the tower?
19.3
Special Right Triangles
There are two special right triangles, the 30  60  90 triangle and the
45  45  90 triangle. What makes these two triangles special is that you can obtain
exact trigonometric values from the triangles. (No rounding off is involved, which leads to an
1
exact answer!) The exact value of sin 45 =
, as you will soon discover.
2
1
•
For example, if you wanted to calculate the value of sine 45 from the fraction
,
2
you would have to divide 1 by 1.4142 (which involves rounding the denominator to
four places) making sin 45 not exact anymore.
Therefore, the exact value for sin 45  =
1
2
=
=
1
2

2
2
2
2
Don't forget to rationalize the denominator.
What are the six exact trigonometric ratios for an angle whose
measure is 45 ?
Mathematics A30
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Lesson 19
Activity 19.31
•
Draw and label an isosceles right-angled triangle whose equal sides are each equal
to one unit. (Note: One unit does not necessarily mean 1 cm)
•
Calculate the length of the hypotenuse in the above diagram.
•
Using the diagram you created, solve the six trigonometric ratios.
sin 45 =
opp
1
=
hyp
2
csc 45 =
cos 45 =
sec 45 =
tan 45 =
cot 45 =
Mathematics A30
545
Lesson 19
What are the six exact trigonometric ratios for and angle whose measure is 30
and for an angle whose measure is 60 ?
Activity 19.32
•
Draw an equilateral triangle whose sides are each equal to two units.
•
From one vertex, draw a perpendicular to the base.
(Hint: This perpendicular should produce two 30  60  90 triangles. The
perpendicular should also bisect the base of the equilateral triangle.)
•
Calculate the length of the perpendicular.
•
Using the diagram you created, solve the six trigonometric ratios.
sin 30 =
sin 60 =
cos 30 =
cos 60 =
Mathematics A30
546
Lesson 19
•
tan 30 =
tan 60 =
csc 30 =
csc 60 =
sec 30 =
sec 60 =
cot 30 =
cot 60 =
Many problems in the assignment call for the exact answer only. For a quick
reference chart, fill in the blanks. You may need to go back to Lesson 17 and Lesson
18 to refresh your memory on quadrantal angles.
0
30
45
60
90
Sine
Cosine
Tangent
Cosecant
Secant
Cotangent
It is important to be able to readily calculate these values. Rather than memorize the
values, sketch and label the required triangles each time it is required to find the value of
a trigonometric function of 30, 45 or 60 . After some practice this procedure will be most
efficient since you will be able to quickly sketch or just visualize the sketch.
Mathematics A30
547
Lesson 19
Example 1
Calculate the exact lengths of the missing sides in each of the following
triangles.
A)
B)
Solution:
A)
Sketch the 30  60  90 triangle.
To solve for x, use the tangent ratio.
opp
adj
x
tan 60  =
5
x = tan 60  5 
tan 60  =
x=
 3 5 
x= 5 3
Mathematics A30
548
Lesson 19
To solve for y, use the cosine ratio.
cos 60  =
cos 60  =
y=
y=
y=
y=
B)
adj
hyp
5
y
5
cos 60 
5
1
2
2
5 
1 
10
The slashes indicate an isosceles triangle.
Sketch the 45   45   90  triangle.
To solve for x, use the sine ratio.
opp
hyp
x
sin 45  =
12
x = sin 45 12 
sin 45  =
 2
12 
x = 

 2 
x= 6 2
The value for y will be the same.
y= 6 2
Example 2
Evaluate:
A)
B)
C)
Mathematics A30
tan 30  + 2 sin 30 
cos 45   tan 60   csc 30 
cot 2 60  + sec 2 30 
549
Lesson 19
Solution:
A)
tan 30  + 2 sin 30 
1
1

+ 2 
3
2
1

+ 1
3
3
3
+
3
3
3 3

3

B)
cos 45   tan 60   csc 30 
1
3
2
=


1
1
2
=
2

2
2 3
=
6
=
C)
3  2
2
NOTE: cot 2 30  = cot 30 
cot 2 60   sec 2 30 
2
=
=
=
=
 1 
 2 

 + 

 3
 3
1
4
+
3
3
5
3
2
1
3
2
The value of a trigonometric function of angle  is equal to the value of the same
trigonometric function of the reference angle of  with the appropriate , 
sign depending on the quadrant of the terminal side of angle .
Mathematics A30
550
Lesson 19
Example 4
Express sin  225 as a function of the reference angle and evaluate. Do the
same for tan  225 . Give answers as exact values.
Solution:
y
* Reference angle  45°
* Sine is positive in Quad II.
45°
x
sin  225   sin 45
-225°


y
1
2
2
2
* Reference angle  45°
* Tangent is negative in Quad II.
45°
x
-225°
tan  225   tan 45
 1
Exercise 19.3
1.
Calculate the exact lengths of the missing sides in each of the following cases.
a)
Mathematics A30
b)
551
Lesson 19
c)
d)
e)
f)
14 2
2.
Evaluate each of the following. Be sure to express the answer in simplest
rationalized form and in exact form. (Exact form means radical form. The radicals
are not to be expressed as rounded off decimals.)
a)
cos 60  sin 90  cos 0
b)
sin 90 cos 60  cos 90 sin 60  cos 60
c)
cot 2 60  sec 2 30
d)
sin 45 cot 30 sin 30
e)
cos 30 sin 60  sin 30 cos 60
f)
cos 2 60 tan 2 45  csc 2 60 cos 2 0
g)
sin 30 sin 45 sin 60 cos 30
h)
sin 2 60  cos 2 60
i)
cot 90  cos 0  sin 32 sin 0
j)
sin 3 30  cos3 30
Mathematics A30
552
Lesson 19
3.
Evaluate each of the following by using the same procedure as in Example 4. Give
the exact values — do not use tables or calculators.
a)
sin ( 135)
b)
cos 240
c)
tan 405
d)
csc 450
e)
sec 540
f)
cot ( 210)
g)
sin ( 270)
h)
cos 300
i)
tan 750 
Mathematics A30
553
Lesson 19
Mathematics A30
554
Lesson 19
Answers to Exercises
Exercise 19.1
1.
By observation, we know:
sine is positive in Quadrant I
* When sin 35   0.9918 your calculator is in RADIAN mode.
It should be in DEGREE mode.
2.
a)
sin   0.7335
 47 .18
b)
cos  =  0.3882
 = 112.84°
c)
tan  = 3.247
 = 72.88° + 180°
 = 252.88°
Mathematics A30
555
Lesson 19
d)
sin  =  0 .4529
  26 .93
ref. angle = 26.93°
for 4th Quad    360   26 .93 
 333.07
e)
cos  =  0.6525
 = 130.73°
* not in Quad III
ref. angle 49.27°
 = 180° + 49.27°
= 229.27°
f)
tan  = 1.100
 = 47.73°
Mathematics A30
556
Lesson 19
3.
a)
cos  = 0.7660
 will be in
Quad I
Quad IV
y
 = 40°
 =  40 or 320°
b)
40o
x
40o reference
angle
csc    1.9875
negative in Quad III and Quad IV
1
  1.9875
sin 
1
  1.9875
sin 
( 1.9875)(sin )  1
csc  =
1
 1.9875
sin    0.5031
sin  
y
o
210.21
o
Mathematics A30
x
o
–30.21
30.21
557
 =  30.21
(or 329.79°)
 = 210.21°
Lesson 19
c)
cot   2 .3548 
negative
in Quad II
and Quad IV
cot  
1
 2 .3548
tan 
1
 2 .3548
tan   0 .4247
tan  
  23 .01 
(or 336 .99 )
  180   23 .01 
  156 .99 
d)
sec = 1.8182
positive in Quad I and Quad IV
sec  
1
 1 .8182
cos 
1
1 .8182
cos   0 .5500
cos  
  56 .63 
  56 .63 
(or 303 .37 )
Mathematics A30
558
Lesson 19
e)
sin  =  0 .3486
y
x
–20.40
 =  20 .40 
(or 339.60°)
y

x
20.4o
 = 180° + 20.40°
= 200.40°
f)
sin  = 2.3500
 = no solution
• check the range of sin values

sin 
0°
Conclusion:
 1  sin   1
90°
180°
360°
 90 
 180 
 360 
Mathematics A30
559
Lesson 19
4.
a)
b)
A = 30°
A =  30
}
Particular Solutions
A =  30  n360, n  I
}
General Solution
A = 90°
A = 90° + n360°
c)
A = 180° + 42° = 222°
A =  42 or 318°
A = 222° + n360°, n  I
A = 318  n360 , n  I
d)
A = 0°, 180°
A = n360°
A = 180 + n360°, n  I
or
A = n180°, n  I
e)
(This combines both solutions.)
A = 135°, 225°
A = 135° + n360°, n  I
A = 225° + n360°, n  I
Exercise 19.2
1.
a)
2 2 .6 °
13
12
6 7 .4 °
5
Mathematics A30
560
Lesson 19
b)
2.
small triangle
opp
18.9
opp  12.7482
tan 34 
22.1482
18.9
tan   1.1719
tan  
  49.5

x    34
 49.5  34
 15.5
Mathematics A30
561
Lesson 19
3.
height of kite above ground
72 2  h 2  100 2
h 2  4816
h  69.3974
69.3974 + 1.4 = 70.7974

add the height Jodi is holding the kite
The height of the kite is 70.8 m.
Angle cord makes with ground.
• parallel lines cut by a
transversal
• corresponding angles
are congruent
From the diagram . . .
72
100
cos   0.72
cos  
  43.9
Mathematics A30
562
Lesson 19
transversal (cord line)

parallel lines

Because we know corresponding angles are congruent,
the angle the cord makes with the ground is 43.9°
4.
x
1.5 m = 150 cm
60
20 cm
x
150
x  129.9 cm
sin 60 
The maximum efficiency is obtained when the handle is held
129.9 + 20 = 149.9 cm or 1.5 m above the ground.
Mathematics A30
563
Lesson 19
5.
200
y
200
200
y

 747 m
tan 15 0.2679
200
tan 10 
xy
200
200
xy

 1134 m
tan 10 0.1763
tan 15 
The fires are 1134  747  387m apart and the nearest fire is 747
m away.
Exercise 19.3
1.
a)
3 8

2
y
x
8
1
x

3 8
3 y  16
3x  8
sin 60 
30°
2
3
y
60°
8
y
tan 30 
16
3
16 3
y
3
Mathematics A30
564
x
x
8
3
8 3
3
Lesson 19
b)
x 7
sin 45 
7
y
2 7

2
y
2 y  14
y
2.
14
2
c)
1
13 3
x 6 , y
2
2
d)
x  4 2, y  8
e)
x  3 3, y  9
f)
x  14, y  14
a)
1
1
1 1  2
2
2
b)
1 1   0  (sin 60)  1
c)
 1   2 
1 4
5
2

  
   
1
3 3
3
3
 3  3
d)
 1   3  1 
   3 

 

 2   1  2 2 2
e)
 3   3  1  1 
3 1
4



 2   2    2   2   4  4  4  1



f)
 2 
1 4
3 16 19
7
1 
2
 1   


1
  1  
4 3 12 12 12
12
2
 3
g)
 1   2   3   3  3 2

 
16
 2   2   2   2 
h)
 3   1 2 3 1
4


 2    2   4  4  4  1


i)
0  1  sin 32 (0)  1
2
14 2
2
7 2
y
2
2

1 1
 1
2 2
2
3 2
6

4
4
2
2
2
Mathematics A30
565
Lesson 19
3
3.
j)
3
1 3 3
13 3
 1   3 
 

  

8
8
8
2  2 
a)
 2
2
b)
1
2
c)
1
d)
csc 450°
450
360
csc 450   csc 90 
csc 90  
90
1
?
sin 90 
1
1
csc 450   1
?
e)
1
f)
 3
g)
1
h)
1
2
i)
tan 750°
3 60
 2
720
tan 750   tan 30 
* tan is  in QI

1
3

3
3
750  720  30 
Mathematics A30
566
Lesson 19
Mathematics A30
Module 2
Assignment 19
Mathematics A30
567
Lesson 19
5. Staple the completed barcode
sheet on top of this address sheet
(upper left corner.)
4. Staple this sheet to the
appropriately-numbered
assignment. Use one address
sheet for each assignment.
3. Complete the details
in this address box.
2. Number all the pages and place
them in order.
1. Write your name and address and
the course name and assignment
number in the upper right corner of
the first page of each assignment.
Before submitting your
assignment, please complete
the following procedures:
Postal Code:
City/Town, Province
Street Address or P.O. Box
Name
Country
Print your name and address, with postal code. This address sheet
will be used when mailing back your corrected assignment.
Assignment Number
19
Mark Assigned:
Distance-Learning Teacher’s Name
Course Title
Mathematics A30
Course Number
8404
Student Number
Staple here to the
upper left corner of
your assignment
Assignment 19
Values
(40)
A.
Multiple Choice: Select the correct answer for each of the following and place
a check () beside it.
1.
The one expression which is not equal to 0.6428 is ***.
____
____
____
____
2.
Mathematics A30
a.
b.
c.
d.
70°,
70°,
70°,
70°,
 70
290°
110°
250°
If sin A =  0.8660 and tan A =  1.7321 , then A could be ***.
____
____
____
____
4.
cos 50°
cos  50
sin 140
sin  220
If cos A = 0.3420, then the principal values of A to the nearest degree
are ***.
____
____
____
____
3.
a.
b.
c.
d.
a.
b.
c.
d.
600°
840°
1000°
1020°
The principal solutions to tan 1 1 are ***.
____
____
a.
b.
____
c.
____
d.
45°, 225°
45°,  225
1
1
,
45 225
45°,  45
571
Assignment 19
5.
Cos 1
a.
b.
A
B
____
c.
____
d.
c
b
a
c
____
____
6.
7.
8.
Mathematics A30
b
is identical to ***.
c
A
c
B
b
C
a
The equation sec A = x is equivalent to the equation ***.
A = cos x
1
A=
cos x
____
a.
____
b.
____
c.
A
____
d.
A  sec 1 x
1
sec x
The particular solutions to the nearest degree for the equation
csc A = 2 are ***.
____
____
____
a.
b.
c.
____
d.
30°, 330°
30°, 150°
45°, 135°
1
1
,
45  135 
The value of sin  is ***.
____
a.
____
b.
____
c.
____
d.
3
3
3
1
2
3
2
P ( 3 , 1)

572
Assignment 19
9.
10.
The value of cos (180 + ) is ***.
____
a.
____
b.
____
c.
____
d.
3
5
4

5
3
5
4
5

y
(3, 4)

x
B
The length of BC is ***.
____
____
____
____
a.
b.
c.
d.
10 3
45
3
15
A
60
o
C
5 3
11.
12.
The length of BC is ***.
____
a.
____
b.
____
c.
____
d.
B
7
A
45°
C
Sec 2 45° is the same as ***.
____
____
____
____
Mathematics A30
5
7 2
2
7 2
7
2
a.
b.
c.
d.
sec (45 × 45)
2 sec 45
(sec 45) (sec 45)
sec 90
573
Assignment 19
13.
14.
The expression cot 30° + 2 csc 30° has the value ***.
____
a.
____
b.
____
____
c.
d.
16.
1
4
3
3
4 3
1
a.
b.
c.
d.
50°
40°
90°
130°
The value of cos  is ***.
____
____
a.
b.
____
c.
____
d.
y
P (-2, 2)
 2
1
2

2
2
2
x
Cos 30° is equal to ***.
____
____
____
____
Mathematics A30
3
In the diagram, the angle of depression is ***.
____
____
____
____
15.
1
a.
b.
c.
d.
sin 60°
sec 60°
csc 60°
sec 30°
574
Assignment 19
17.
Sec 495° expressed in terms of the reference angle is ***.
____
____
____
____
18.
20.
Cot
a.
b.
c.
d.
0
1
1
undefined
7
7
 , where  is radian measure, has the value ***.
3
3
____
a.
____
b.
____
c.
____
d.
1

3
1
3
1
2
2
The value of cos 3  , where 3  is radian measure, is ***.
____
____
____
____
Mathematics A30
sec 135°
sec 45°
 sec 135
 sec 45
Csc 180° is ***.
____
____
____
____
19.
a.
b.
c.
d.
a.
b.
c.
d.
0
1
1
undefined
575
Assignment 19
Mathematics A30
576
Assignment 19
Answer Part B and Part C in the space provided. Evaluation of your solution to each
problem will be based on the following.
•
A correct mathematical method for solving the problem is shown.
•
The final answer is accurate and a check of the answer is shown where asked for by
the question.
•
The solution is written in a style that is clear, logical, well organized, uses proper
terms, and states a conclusion.
(8)
B.
(8)
1.
Solve the triangle if B is 18.2° and c is 43.1 m. Give lengths to one
decimal place and angles to the nearest tenth of a degree.
2.
One end of 30.4 m long cable is attached to the top of a 22 m tall
vertical pole. The cable is extended as far as possible from the base of
the pole and is attached to the ground.
Mathematics A30
a)
What angle, to the nearest degree, does the cable make with the
ground?
b)
How high up the pole is the cable a horizontal distance of one
meter from the pole?
577
Assignment 19
(8)
3.
Solve triangle ABC giving angles to the nearest tenth of a degree and
lengths to one decimal place.
B
9
A
Mathematics A30
12
C
578
7
D
Assignment 19
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4.
Mathematics A30
From the top of a 500 m tall building the angle of depression to the top
of another smaller building is 40° and to the base of the smaller
building is 50°.
a)
Find the horizontal distance between the buildings.
b)
Find the height of the smaller building.
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Assignment 19
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5.
At a certain distance away from
the base of a tower, the angle of
elevation to the top is 32° and
122 m further away from the
base the angle of elevation is 22°.
Find the height of the tower.
t ower
122 m
Mathematics A30
580
Assignment 19
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C.
1.
Mathematics A30
For a to g, verify that each statement is true by evaluating the terms
on both sides of the equal sign and showing that the values are equal.
(Use exact values. No calculators.)
a.
sin 2 60  cos 2 60  sin 2 45  cos 2 45
b.
1  2 sin 2 30  cos 60
c.
2 cos 2 30  1  cos 60
d.
sin 45 
1  cos 90
2
e.
cos 90 
1  cos 180
2
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Assignment 19
Mathematics A30
f.
tan 60  tan 30
 tan 30
1  tan 60 tan 30
g.
1  cot 30 cot 60
 cot 30
cot 30  cot 60
h.
Evaluate
sin 390° + cos ( 45)  sin ( 225) .
i.
Evaluate
sin 0° + cos 180°  sin 270 .
j.
Evaluate
sin 2


5
 cos 2  sin 2
.
3
6
3
582
Assignment 19
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2.
(STUDENT JOURNAL)
Write a summary of the material in this lesson which will be useful for
study and review.
100
Mathematics A30
583
Assignment 19
Mathematics A30
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Assignment 19