Download Pre-Class Problems 9

Pre-Class Problems 9 for Friday, February 24
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
Find the exact value of  , x, and z. Then approximate the value of x
and z to the nearest tenth.
b.
49.3
z
x

24.2
Additional problems available in the textbook: Page 494 … 51 – 54.
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
7
cannot 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 cannot
take the inverse sine of numbers greater than one.
Back to Problem 1.
2a.
x

y
34.7
26.8
To find  :
  26.8   90     63.2 
Answer: 63.2 
To find x:
x
 sin 26.8   x  34.7 sin 26.8 
34.7
Answer:
Exact: x  34.7 sin 26.8 
Approximate: 15.65
NOTE: sin 26.8   0 . 4508775407
To find y:
y
 cos 26.8   y  34.7 cos 26.8 
34.7
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
OR
24.2
24.2
 tan 49.3   24.2  x tan 49.3   x 
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
24.2
24.2

sin
49
.
3


24
.
2

x
sin
49
.
3


x

OR
z
sin 49.3 
Answer:
24.2
z

z

24
.
2
csc
49
.
3

Exact:
OR
sin 49.3 
Approximate: 31.9
NOTE: sin 49.3   0 . 7581343362
csc 49.3   1.31902745
Back to Problem 2.
Solution to Problems on the Pre-Exam:
13.
Back to Page 1.
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 