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Pre-Class Problems 8 for Tuesday, February 25
These are the type of problems that you will be working on in class. These
problems are from Lesson 7.
Solution to Problems on the Pre-Exam.
You can go to the solution for each problem by clicking on the problem letter.
Objective of the following problems: To use a calculator to approximate the value
of a trigonometric function of an angle.
1.
Use your calculator to approximate the following to four decimal places.
(Round to the nearest ten-thousandth.)
a. sin (  215 )
d.
 26  
cos  

11 

b.
sec
9
7
c.
cot 289 
e.
tan 1890.4 
f.
csc 14
Objective of the following problems: To solve for unknowns in a given right
triangle. To use a calculator to obtain approximations for the exact answers.
2.
Solve for the following variables.
a.
Find the exact value of  , x, and y. Then approximate the value of x
and y to the nearest hundredth.
x

y
26.8
34.7
b.
Find the exact value of  , x, and z. Then approximate the value of x
and z to the nearest tenth.
49.3
z
x

24.2
Objective of the following problems: To take a written description and produce a
right triangle with known information and one unknown. The unknown
information is represented by a variable. Then use a trigonometric function to
obtain an equation containing the variable. Solve this equation for the exact value
of the variable. Then approximate the exact value of the variable as indicated.
For some of these problems, you will need the definition for angle of elevation and
for angle of depression.
An angle of elevation and an angle of depression are both acute angles measured
with respect to the horizontal. An angle of elevation is measured upward and an
angle of depression is measured downward. The angle  below is an angle of
elevation from the point A to the point B above. The angle  below is an angle of
depression from the point B to the point A below.


A
B
B


A
3a.
The angle of depression from the top of a building to an object on the ground
is 40 . If the object is 85 feet from the base of the building, then find the
height of the building. Find the exact value and then round to the nearest
tenth.
3b.
The angle of depression from the top of a 150-foot building to an object on
the ground is 24.7 . How far is the object from the base of the building?
Find the exact value and then round to the nearest hundredth.
3c.
From a point P on the ground, the angle of elevation to the top of a 60-yard
tree is 35 . What is the distance from the point P to the top of the tree? Find
the exact value and then round to the nearest hundredth.
3d.
The angle of elevation of the string from the ground to a kite is 48.6 . If the
length of the string is 125 meters, then how far is the kite above ground?
Find the exact value and then round to the nearest tenth.
3e.
An observer on the ground is 105 yards from the point directly beneath a
balloon. If the angle of elevation from the observer to the balloon is 28 ,
then how far is the balloon from the observer? Find the exact value and then
round to the nearest hundredth.
3f.
A ladder is leaning against the top of a vertical wall. The top of the ladder
makes an angle of 34 with the wall. If the height of the wall is 6 meters,
then find the length of the ladder. Find the exact value and then round to the
nearest tenth.
3g.
The angle of elevation from an object on the ground to the top of a building is
57 . If the object is 95 meters from the top of the building, then find the
distance from the object to the base of the building. Find the exact value and
then round to the nearest thousandth.
3h.
From a point on the ground which is 40 feet from the base of a tree, the angle
of elevation to the top of the tree is 72.3 . What is the height of the tree?
Find the exact value and then round to the nearest tenth.
3i.
A ladder is leaning against the top of a 15-yard vertical wall. The bottom of
the ladder makes an angle of 24.1 with the ground. How far is the bottom
of the ladder from the base of the wall? Find the exact value and then round
to the nearest hundredth.
Additional problems available in the textbook: Examples 8, 9, and 10 on page 144.
Page 147 … 63, 64, 65, 66, 67, 70, 71. Page 196 … 5, 6, 7, 8, 19, 20, 21, 22.
Examples 1 and 2 on Page 190.
Requires a system of equations to solve: Page 148 … 72. Page 196 … 23, 24, 25.
For Problem 23, use 47.67  for 47  40 . Example 3 on Page 191.
Solutions:
1 a.
sin (  215 )
Answer: 0.5736
NOTE: In order to find the sine of the angle  215  , the mode of your
calculator needs to be set on Degrees. If your calculator is set on Radians,
then you would incorrectly give an answer of  0 . 9802 . Since you know
that the terminal side of the angle  215  is in the second quadrant, where
sine is positive, then you would know that this value is not correct.
Back to Problem 1.
1 b.
sec
9
7
Answer:  1. 6039
NOTE: The secondary key of COS  1 , which is above the COS key, on
your calculator is NOT the secant key. It is the key for the inverse cosine
function which we will study in Lesson 9.
NOTE: Since your calculator does not have a secant key, you will first need
9
to find the cosine of the angle
. Do not round this number, which is
7
 0 . 6234898019 . Now, find the multiplicative inverse (reciprocal) of this
1
number using your reciprocal key, which is x
or 1 / x , in order to obtain
9
the secant of the angle
since secant is the reciprocal of cosine.
7
9
NOTE: In order to find the cosine of the angle
, the mode of your
7
calculator needs to be set on Radians. If your calculator is set on Degrees,
9
then you would incorrectly give an answer of 1.0025 for sec
. Since
7
9
you know that the terminal side of the angle
is in the third quadrant,
7
where secant is negative, then you would know that this value is not correct.
If the mode of your calculator was set on Radians and you used the secondary
key of COS  1 , then your calculator would give you an error message since
9
the number
is greater than one. We will learn in Lesson 9 that you can
7
not take the inverse cosine of numbers greater than one.
Back to Problem 1.
1c.
cot 289 
Answer:  0 . 3443
NOTE: The secondary key of TAN  1 , which is above the TAN key, on
your calculator is NOT the cotangent key. It is the key for the inverse
tangent function which we will study in Lesson 9.
NOTE: Since your calculator does not have a cotangent key, you will first
need to find the tangent of the angle 289  . Do not round this number, which
is  2 . 904210878 . Now, find the multiplicative inverse (reciprocal) of this
1
number using your reciprocal key, which is x
or 1 / x , in order to obtain
the cotangent of the angle 289  since cotangent is the reciprocal of tangent.
NOTE: In order to find the tangent of the angle 289  , the mode of your
calculator needs to be set on Degrees. If your calculator is set on Radians,
then you would incorrectly give an answer of  37 . 6927 for cot 289  . If
the mode of your calculator was set on Degrees and you used the secondary
key of TAN  1 , then you would incorrectly give an answer of 89.8017
cot 289  . Since you know that the terminal side of the angle 289  is in the
fourth quadrant, where cotangent is negative, then you would know that this
value is not correct.
Back to Problem 1.
1d.
 26  
cos  

11 

Answer: 0.4154
26 
, the mode of your
11
calculator needs to be set on Radians. If your calculator is set on Degrees,
then you would incorrectly give an answer of 0.9916.
NOTE: In order to find the cosine of the angle 
Back to Problem 1.
1e.
tan 1890.4 
Answer:  143. 2371
NOTE: In order to find the tangent of the angle 1890.4  , the mode of your
calculator needs to be set on Degrees. If your calculator is set on Radians,
then you would incorrectly give an answer of  1.1129 .
Back to Problem 1.
1f.
csc 14
Answer: 1.0095
NOTE: The secondary key of SIN  1 , which is above the SIN key, on your
calculator is NOT the cosecant key. It is the key for the inverse sine function
which we will study in Lesson 9.
NOTE: Since your calculator does not have a cosecant key, you will first
need to find the sine of the angle 14 (radians). Do not round this number,
which is 0.9906073557. Now, find the multiplicative inverse (reciprocal) of
1
this number using your reciprocal key, which is x
or 1 / x , in order to
obtain the cosecant of the angle 14 (radians) since cosecant is the reciprocal
of sine.
NOTE: In order to find the sine of the angle 14 (radians), the mode of your
calculator needs to be set on Radians. If your calculator is set on Degrees,
then you would incorrectly give an answer of 4.1336 for csc 14 . If the mode
of your calculator was set on Radians and you used the secondary key of
SIN  1 , then your calculator would give you an error message since the
number 14 is greater than one. We will learn in Lesson 9 that you can not
take the inverse sine of numbers greater than one.
Back to Problem 1.
2a.
x

y
26.8
34.7
To find  :
  26.8   90     63.2 
Answer: 63.2 
x
 sin 26.8   x  34.7 sin 26.8 
34.7
To find x:
Answer:
Exact: x  34.7 sin 26.8 
Approximate: 15.65
NOTE: sin 26.8   0 . 4508775407
y
 cos 26.8   y  34.7 cos 26.8 
34.7
To find y:
Answer:
Exact: y  34.7 cos 26.8 
Approximate: 30.97
NOTE: cos 26.8   0 .8925858185
Back to Problem 2.
2b.
49.3
z
x

24.2
To find  :
  49.3   90     40.7 
Answer: 40.7 
To find x:
24.2
x
 tan 49.3  
 cot 49.3   x  24.2 cot 49.3 
x
24.2
24.2
24.2

tan
49
.
3


24
.
2

x
tan
49
.
3


x

OR
x
tan 49.3 
Answer:
24.2
x

Exact: x  24.2 cot 49.3  OR
tan 49.3 
Approximate: 20.8
NOTE: tan 49.3   1.162607256
cot 49.3   0 .8601356946
To find z:
24.2
z
 sin 49.3  
 csc 49.3   z  24.2 csc 49.3 
z
24.2
OR
24.2
24.2
 sin 49.3   24.2  x sin 49.3   x 
z
sin 49.3 
Answer:
24.2
z

Exact: z  24.2 csc 49.3  OR
sin 49.3 
Approximate: 31.9
NOTE: sin 49.3   0 . 7581343362
csc 49.3   1.31902745
Back to Problem 2.
3a.
Top of Building
-------------40 
y
40 
Object
85 feet
NOTE: Since the angle of depression is 40  , then the angle of elevation is
also 40  .
y
 tan 40   y  85 tan 40 
85
Answer:
Exact: 85 tan 40  ft
Approximate: 71.3 ft
Back to Problem 3.
3b.
Top of Building -------------24.7 
150 feet
24.7 
x
Object
NOTE: Since the angle of depression is 24.7  , then the angle of elevation is
also 24.7  .
150
x
 tan 24.7 
 cot 24.7  x  150 cot 24.7
x
150
150
150

tan
24
.
7


150

x
tan
24
.
7


x

OR
x
tan 24.7
150
tan 24.7 ft
Exact: 150 cot 24.7 ft OR
Answer:
Approximate: 326.12 ft
Back to Problem 3.
3c.
Top of Tree
z
60 yards
P
35
60
z
 sin 35 
 csc 35  z  60 csc 35
z
60
60
60

sin
35


60

z
sin
35


z

OR
z
sin 35
Answer:
Exact: 60 csc 35 yd OR
60
sin 35 yd
Approximate: 104.61 yd
Back to Problem 3.
3d.
Kite
125 meters
y
48.6
y
 sin 48.6   y  125 sin 48.6 
125
Answer:
Exact: 125 sin 48.6  m
Approximate: 93.8 m
Back to Problem 3.
3e.
Balloon
z
Observer
28
105 yards
105
z
 cos 28 
 sec 28  z  105 sec 28
z
105
OR
105
105
 cos 28  105  z cos 28  z 
z
cos 28
Answer:
Exact: 105 sec 28 yd OR
105
yd
cos 28
Approximate: 118.92 yd
Back to Problem 3.
3f.
Top of Wall
34
z
6 meters
6
z
 cos 34 
 sec 34  z  6 sec 34
z
6
OR
6
6
 cos 34  6  z cos 34  z 
z
cos 34
Answer:
Exact: 6 sec 34 m OR
Approximate: 7.2 m
6
cos 34 m
Back to Problem 3.
3g.
Top of Building
95 meters
57 
Object
x
x
 cos 57  x  95 cos 57
95
Answer:
Exact: 95 cos 57 m
Approximate: 51.741 m
Back to Problem 3.
3h.
Top of Tree
y
72.3
40 feet
y
 tan 72.3  y  40 tan 72.3
40
Answer:
Exact: 40 tan 72.3 ft
Approximate: 125.3 ft
Back to Problem 3.
3i.
Top of Wall
15 yards
24.1
x
15
x
 tan 24.1 
 cot 24.1  x  15 cot 24.1
x
15
OR
15
15
 tan 24.1  15  x tan 24.1  x 
x
tan 24.1
Answer:
Exact: 15 cot 24.1 yd OR
15
tan 24.1 yd
Approximate: 33.53 yd
Back to Problem 3.
Solution to Problems on the Pre-Exam:
Back to Page 1.
13.
Given the triangle below, find x. Set up an equation and solve. (4 pts.)
45
x
28 
45
x
 tan 28  
 cot 28   x  45 cot 28 
x
45
45
45

tan
28


45

x
tan
28


x

OR x
tan 28 
45
x

x

45
cot
28

Answer:
OR
tan 28 
15.
From the top of a building, which is 80 meters tall, the angle of depression to
an object on level ground below is 15.7  . How far is the object from the top
of the building? Draw a picture and label known information. Indicate any
variable you use. Set up an equation and solve. (6 pts.)
Top of Building
-------------15.7 
z
80 meters
15.7 
Object
NOTE: Since the angle of depression is 15.7  , then the angle of elevation is
also 15.7  .
80
z
 sin 15.7 
 csc 15.7  z  80 csc 15.7
z
80
80
80

sin
15
.
7


80

z
sin
15
.
7


z

OR
z
sin 15.7
Answer:
80 csc 15.7 m OR
80
m
sin 15.7