Download Section 7.3 Computing Trigonometric Values of Acute Angles

23
Sect 7.3 – Computing the Values of Trigonometric
Functions of Acute Angles
Objective 0:
Understanding Special Right Triangles.
The first special right triangle we want to examine is a 45˚-45˚-90˚ right
triangle. Since two of the angles are the same, that means two of the sides
are equal, so we have an isosceles triangle. Also, the two equal sides of
the isosceles triangle are the legs of the right triangle since the hypotenuse
is always opposite of the right angle. If we let a be the length of one of the
equal sides, then we can use the Pythagorean Theorem to find the
hypotenuse.
c2 = a2 + a2 (combine like terms)
c2 = 2a2
(take the square root)
c=
a
a 2
45˚
2a2
c=a 2
€
45˚
a
This
€ means that the ratio of sides for a 45˚-45˚-90˚ is a: a: a 2 .
€
The second special right triangle we want to examine is a 30˚-60˚-90˚ right
triangle. If we take an equilateral triangle and cut it in half along the altitude,
we get two 30˚-60˚-90˚ triangles. The shortest side€is half of the longest
side. If we let a be the length of the shortest side, then 2a will be the length
of the longest or the hypotenuse. We can use the Pythagorean Theorem to
find the other leg of the triangle.
(2a)2 = a2 + h2
4a2 = a2 + h2
– a2 = – a2
3a2 = h2
(simplify)
(subtract a2)
h2 = 3a2
(take the square root)
h=
a 3 30˚
3a2 = a 3
€
2a
60˚
a
This means that the ratio of sides for a 30˚-60˚-90˚ is a: a 3 : 2a.
€
€
€
24
Complete the following table:
Ex. 1
θ
θ
(radians) (degrees)
π
6
π
4
π
3
€
sin θ
cos θ
csc θ
sec θ
€
30˚
45˚
60˚
2a
a 3 30˚
60˚
a
€
sec 30˚ =
1
cosθ
=
sin 30˚ =
opp
hyp
=
a
2a
=
1
2
cos 30˚ =
adj
hyp
=
a 3
2a
=
tan 30˚ =
opp
a
=
€
adj
a 3
1
2
=€ =
sinθ
1
=
csc 30˚€=
2
3
•
3
3
2 3
3
=
3
2
sin 60˚ = cos 30˚ =
€
cot €
30˚ =
€
€
1
tanθ
€ 60˚ = sin 30˚ =
cos
csc 60˚€= sec€30˚ =
sec 60˚ = csc
€ 30˚ = 2
cot 60˚ = tan 30˚ =
sin θ
cos θ
tan θ
1
2
3
2
3
3
€1
π
6
30˚
π
4
45˚
π
3
€
60˚
€3
€
€
2
3
2
€
csc θ
€
2
€
2
3
3
3
2
1
3
€
3
3
=
3
3
=€ 3
1
2
2 3
3
3
3
sec θ
cot θ
2 3
3
3
€
2
3
3
€
€
•
2
tan
€ 60˚ =€cot€30˚ = € 3
θ
θ€
(radians) (degrees)
€
cot θ
Solution:
One of our special triangles from before was a 30-60-90 triangle:
€
€
tan θ
€
25
Another of our special triangles was a 45-45-90 triangle:
a
a 2
45˚
45˚
€
1
cosθ
θ
θ
(radians) (degrees)
€
€
π
30˚
6
€
€
€
opp
hyp
=
cos 45˚ =
adj
hyp
=
csc 45˚ =
€
cot 45˚ =
2
=
€
tan θ
€
cos θ
€
sin θ
1
2
3
2
3
3
1
π
4
45˚
2
2
2
2
π
3
€
60˚
€3
€1
a 2
a
=
1
=
1
1
2
2
•
2
•
2
2
2
sec θ
2
2 3
3
3
3
2
2
=
2
2
€
csc θ
€
2
=
€
2
3
2
2
a
a 2
opp
a
=
=
adj € a €
1
= 2
sinθ €
€
1
=1
tanθ
tan 45˚ €
=
a
sec 45˚ =
sin 45˚ =
cot θ
3
1
2
€
2
3
3
€
€
€
€
Find the exact value of each expression:
π
π
€
Ex. 2a sin 30˚€cos 45˚ €
Ex.
– tan€
€ 2b cot
6
Ex. 2c
sin
2
Solution:
a)
€
b)
c)
π
4
– csc
2
sin 30˚ cos 45˚ =
€
π
π
cot
– tan
=
6
π
sin2
4
€ 2:
Objective
6
π
3
6
π
–€
csc€2
3
2
1
•
2
2
3 –
(
=€
2
2
=
2
4
3
3
=
2
) –(
€
€
3 3
3
2 3
–
=
3
3
3
2 3 2
1
4
3
= –
=
2
3
6
3
)
–
8
6
=
5
6
€
€ the€Value of
€ a Trigonometric
€
€ Approximate
Function.
€ SIN key
€ is for the
€
€
€ sine
€ ratio,
€ COS
€ key
€ is for the
On a scientific
calculator,
cosine ratio, and TAN key is for the tangent ratio. To evaluate a
trigonometric function for a specific angle, you first want to be in the correct
26
mode. If the angle is in degree, then the calculator needs to be in degree
mode while if the angle is in radians, the calculator needs to be in radian
mode. Next, enter the trigonometric function first and then the angle. Close
you angle with a ")" and hit enter. For the secant, cosecant, and cotangent,
we want to use the following facts:
1
1
1
sec(θ) =
csc(θ) =
cot(θ) =
cos(θ)
sin(θ)
tan(θ)
Evaluate the following (round to three decimal places):
Ex. 3a
€
Ex. 3c
cos(26˚)
2
sin(16 ˚)
Ex. 3e
sec(1.2)
€
3
Ex. 3b
Ex. 3d
tan(78.3˚)
€
csc(56.11˚)
Ex. 3f
tan(
3π
2
)
Solution:
a)
We put our calculator in degree mode, hit the COS key, type
26) and hit enter:
cos(26˚) = 0.8987… ≈ 0.899
b)
We put our calculator in degree mode, hit the TAN key, type
78.3) and hit enter:
tan(78.3˚) = 4.8289… ≈ 4.829
c)
We put our calculator in degree mode, hit the SIN key, type 16+
2
2/3) and hit enter:
sin(16 ˚) = 0.2868… ≈ 0.287
3
d)
We put our calculator in degree mode, hot the SIN key, type
56.11) and hit enter: sin 56.11˚ = 0.8308…
Now, hit the x – 1 key to take the reciprocal:
1
1
csc(56.11˚) =
=
= 1.2046… ≈ 1.205
sin(56.11)
e)
Since there is no degree mark on the angle, then the angle is in
radians. We put our calculator in radian mode, hit the COS key,
type 1.2) and
cos(1.2) = 0.3623…
€ hit enter:€
–1
Now, hit the x key to take the reciprocal:
1
1
sec(1.2) =
=
= 2.7597… ≈ 2.760
cos(1.2)
f)
0.8308...
0.3623...
Since there is no degree mark on the angle, then the angle is in
radians. We put our calculator in radian mode, hit the TAN key,
€
€
3π
type 3π/2) and hit enter: tan(
) = Error means it is undefined.
2
27
Find the missing sides and angles (to the nearest tenth):
F
B
Ex. 4
Ex. 5
18 ft
A
10.1 m
23˚
C
42.3˚
Solution:
Since ∠A and ∠B are
complimentary, then
m∠B = 90˚ – 23˚ = 67˚.
We have the hypotenuse
and the acute angle A. We
can use the sine of 23˚ to
G
E
Solution:
Since ∠E and ∠G are
complimentary, then
m∠G = 90˚ – 42.3˚ = 47.7˚
The adjacent side to ∠G is 10.1 m
Using the tangent function, we can
find the opposite side EF of ∠G.
find BC and the cosine of
tan(47.7˚) =
23˚ to find AC.
opp
sin(23˚) =
=
To solve, multiply by 18:
BC = 18•sin(23˚)
= 18•0.3907… = 7.03…
≈ 7.0 ft
adj
AC
cos(23˚) =
=
To solve, multiply by 10.1
EF = 10.1•tan(47.7˚)
€
€
= 10.1•1.098985…
= 11.0997… ≈ 11.1 m
To find the length EG, we can
use the cosine function:
adj
10.1
cos(47.7˚) =
=
To solve for AC, multiply
by 18:
AC = 18•cos 23˚
= 18•0.9205… = 16.56…
≈ 16.6 ft
Thus, BC ≈ 7.0 ft,
AC ≈ 16.6 ft, and m∠B = 67˚.
Multiply both sides by EG and
divide both sides by cos(47.7˚):
EG = 10.1/cos(47.7˚)
€
EG = 10.1/0.6730… ≈ 15.0 m
Hence, m∠G = 47.7˚,
EF ≈ 11.1 m, and EG ≈ 15.0 m.
hyp
hyp
Objective 4;
BC
18
18
opp
adj
hyp
=
EF
10.1
EG
Solving Applied Problems.
The key to solving applications involving right triangles is drawing a
reasonable diagram to represent the situation. Next, we need determine
what information we are given and what information we need to find.
Finally, we then use the appropriate trigonometric function to solve the
problem.
28
Solve the following:
Ex. 6
A surveyor measures the angle between her instruments and
the top of a tree and finds the angle of elevation to be 65.7˚. If her
instruments are 4 ft high off of the ground and are 25 feet away from
the center of the base of the tree and the ground is leveled, how high
is the tree? Round to the nearest tenth.
Solution:
We begin by drawing the surveyor's
instruments and the tree. Since her
instruments are 4 ft above the ground,
we can draw a horizontal line from her
her instruments to four feet above the
x
center of the base of the tree. Drawing
a line from the top of the tree to her
instruments and a vertical line down
65.7˚
from the top of the tree four feet above
25 feet
the center of the base of the tree, we
get a right triangle. The adjacent side
is 25 feet and we are looking for the opposite of the triangle. So, we
will need the tangent function. Hence,
opp
x
tan(65.7˚) =
=
(multiply by 25 to solve for x)
adj
25
x = 25•tan(65.7˚) = 25•2.214… = 55.36… ft
Now, add 4 ft to find the height of the tree.
So, the tree is approximately 59.4 ft tall.
Ex. 7
A helicopter flying directly above an ocean liner at a height of
1400 ft, locates another boat at a 23˚ angle of depression. How far
apart are the two ships? Round to the nearest hundred feet.
Solution:
The angle of depression is the angle between a horizontal line and
the line of observation. In other words, it is the angle one has to look
down to see the observed object, which is a ship in this case. The
helicopter is flying directly above the ocean liner. We can draw a
vertical line between the helicopter and the ocean liner. Next we can
draw a line from the helicopter to the other ship. The angle between
these two lines and the angle of depression are complimentary
29
angles. Thus, the angle between
23˚
the two lines is 90˚ – 23˚ = 67˚.
67˚
Now, drawing a line between the
two ships creates a right triangle. 1400 ft
The vertical height is the adjacent
side of 67˚ and we are looking for
the opposite side of 67˚. So, we will
need to use the tangent function.
x
opp
x
tan(67˚) =
=
(multiply by 1400 to solve for x)
adj
1400
x = 1400•tan(67˚) = 1400•2.35… = 3298.1… ft ≈ 3300 ft
So, the ships are approximately 3300 ft apart.
Ex. 8
The distance from a water source that is two feet below the
ground to the input of a tank 63 ft. If the angle of elevation is 12.3˚
from the source to the input of the tank, how much pipeline (to the
nearest tenth) will be needed if the plumber runs the pipe horizontally
two feet under the ground from the water source to right below the
input of tank and then turn the pipe vertically and run it to the input?
Solution:
We begin by drawing the tank and
the water source. We draw a line
from the input of the tank to the
63 ft
water source. We then draw two
y
lines for where the pipe will run.
The two lines for the pipe are the
legs of a right triangle and the 63
12.3˚
ft is the length of the hypotenuse.
The angle of elevation is the acute
x
angle we will be using. To find the
opposite side (y), we will need to use the sine function and to find the
adjacent side, we will need to use the cosine function.
opp
y
sin(12.3˚) =
=
(multiply by 63 to solve for y)
hyp
63
y = 63•sin(12.3˚) = 63•0.213… = 13.42… ft
cos(12.3˚) =
adj
hyp
=
x
63
(multiply by 63 to solve for x)
x = 63•cos(12.3˚) = 63•0.977… = 61.55… ft
Now, add the two results: 61.55… + 13.42… = 74.97… ≈ 75.0 ft
The plumber will need 75.0 ft of pipe.
30
Ex. 9 Two cities, San Antonio and Philadelphia, are 1494 miles apart.
To avoid a bad storm, a pilot took a course that was 20˚ further
south from San Antonio for 400 miles and then turned to fly towards
Philadelphia. How much further did the pilot have to fly to reach
Philadelphia? Round to the nearest mile.
Solution:
If we draw the picture, we realize that we
Philadelphia
may not have a right triangle
since when the pilot
1494 miles
turned the plane, there
?
is no guarantee that
he or she turned at
20˚
a 90˚ angle. Let's
SA
400 miles
suppose that the pilot
had went further and then turned at a 90˚.
We can use the cosine function to find the adjacent side (x):
adj
x
Philadelphia
cos 20˚ =
=

hyp
1494
(multiply by 1494)
1494 miles
x = 1494•cos 20˚
y
?
x = 1494•0.939…
SA
x = 1403.9… mi
20˚
The extra distance the
400 miles
pilot would have had to
x
travel in order to turn at a 90˚ is 1403.9… – 400 = 1003.9… miles.
Now use the Pythagorean Theorem to find the opposite side (y):
y2 + (1403.9…)2 = (1494)2
y2 + 1970937… = 2232036 (subtract 1970937…)
y2 = 261098…
y = 261098... = 510.97… mi
?
Now consider the smaller right triangle
510.9…miles
in the diagram. We know both legs so we
can use the Pythagorean Theorem to find ?
?2 = (1003.9…)2 + (510.9…)2
1003.9…miles
?2 = 1007816…+ 261098… = 1268915.38
? = 1268915.38 = 1126.4…
So, the plane traveled 400 + 1126.4… = 1526.4… miles.
Thus, 1526.4 – 1494 = 32. 46… ≈ 32 miles
So, the plane traveled 32 miles further.