Name SOLUTIONS Section

Rec. Instr.

PLANE TRIGONOMETRY

Exam IIINovember 15, 2001

Below you will find 10 problems, each worth 10 points. Solve the problems in the spaceprovided (use the back page as scratch paper). When writing a solution to a problem, showall work. No books or notes are allowed. Sign and submit your formula sheet withthe exam.

Problem Score

1

2

3

4

5

Problem Score

6

7

8

9

10

Total (out of 100 points):

Problem 1. Find the exact value of arccos

(−√

3

2

).

arccos

(−√

3

2

)= π − arccos

(√3

2

)= π − π

6=

5π

6.

Problem 2. Compute:

(a) cos(arctan 34).

Denote arctan 34

by t, so that tan t = 34, and −π

2< t < π

2. Since tan t is positive, it follows

that t is in the first quadrant. This gives

sec t =√

1 + tan2 t =

√1 +

(3

4

)2

=

√1 +

9

16=

√25

16=

5

4,

so we get cos t = 45.

(b) tan(2 arctan 34).

tan 2t =2 tan t

1− tan2 t=

2 · 3

4

1−(

3

4

)2 =

3

2

1− 9

16

=

3

27

16

=3

2· 16

7=

24

7.

Problem 3. A triangle ABC has c = 20 in, A = 60◦ and B = 45◦. Approximate theremaining parts. (Express all angles in degrees. Round to two decimal places.)

C = 180◦−A−B = 180◦−60◦−45◦ = 75◦. Using the Law of Sinesa

sin A=

b

sin B=

c

sin Cwe have

a

sin 60◦=

b

sin 45◦=

20

sin 75◦.

The equalitya

sin 60◦=

20

sin 75◦gives

a =20 sin 60◦

sin 75◦' 17.93,

and the equalityb

sin 45◦=

20

sin 75◦gives

b =20 sin 45◦

sin 75◦' 14.64.

1

Problem 4. Find the equation of the line which passes through the point P (7,−3) and isperpendicular to the line x− 2y = 8.

The given line has the equation x− 8 = 2y, which can be written as y = 12(x− 8), that is:

y =1

2x− 4.

This means that the given line has slope m1 = 12.

Since the unknown line is perpendicular to the given line, its slope m must satisfy m·m1 =−1, so we get

m = − 1

m1

= −2.

Using the Point-Slope Equation y = m(x − x1) + y1, the unknown line has the equationy = −2(x− 7)− 3, which is:

y = −2x + 11.

Problem 5. Find the equation of the parabola which has vertex V (2, 1) and focus F (3, 1).

Observe first that the symmetry line V F is horizontal. This means that the equation ofthe parabola is of the form

4p(x− h) = (y − k)2,

where (h, k) are the coordinates of the vertex, and (h + p, k) are the coordinates of the focus.In our case we get h = 2, k = 1, and 2 + p = 3, which gives p = 1. So the equation of theparabola is:

4(x− 2) = (y − 1)2.

Problem 6. Write the complex number 2− 2i in trigonometric form: r(cos θ + i sin θ) with0 ≤ θ < 2π.

Writing our number as a + bi we have a = 2 and b = −2. Then

r =√

a2 + b2 =√

22 + (−2)2 =√

8 = 2√

2.

The angle θ is then determined by

cos θ =a

r=

2

2√

2=

√2

2and sin θ =

b

r=−2

2√

2= −

√2

2.

It is clear that θ =7π

4satisfies the above equalities. So the trigonometric form is

2− 2i = 2√

2(cos

7π

4+ i sin

7π

4

).

2

Problem 7. A triangle ABC has b = 30 in, A = 60◦ and c = 20 in.(a) Approximate the remaining parts. (Express all angles in degrees. Round to two decimalplaces.)

First we use the Law of Cosine to find the side a:

a =√

b2 + c2 − 2bc cos A =√

302 + 202 − 2 · 30 · 20 · 12

=√

700 = 10√

7 ' 26.46.

Next we find the angle B using the Law of Cosine:

B =180◦

π· cos−1

(a2 + c2 − b2

2ac

)=

180◦

π· cos−1

(700 + 202 − 302

2 · 10√

7 · 20

)' 79.11◦.

Finally, we findC = 180◦ − A− B ' 40.89◦.

(b) Find the area of the triangle.

We use the formula

A =1

2bc sin A =

1

2· 30 · 20 ·

√3

2= 150

√3 ' 259.81.

Problem 8. A straight line passes through the points A(3, 2) and B(1, 6).

(a) Find the slope of the line.

Use the formula for the slope:

m =y1 − y2

x1 − x2

=2− 6

3− 1=−4

2= −2.

(b) Find the equation of the line in the form y = mx + b.

Using the Point-Slope Equation y = m(x − x1) + y1, the line has the equation y =(−2)(x− 3) + 2, which is:

y = −2x + 8.

(c) Find the x-intercept and the y-intercept.

The above equation already gives the y-intercept: (0, 8). To find the x-intercept, we sety = 0 and we solve for x. We have −2x + 8 = 0, which gives 8 = 2x, thus x = 4. Thex-intercept is the point (4, 0).

3

Problem 9. A triangle ABC has b = 100 in, B = 30◦ and c = 141 in. Approximate theremaining parts. (Express all angles in degrees. Round to two decimal places.)

Using the Law of Sines, we have

a

sin A=

100

sin 30◦=

141

sin C.

Using the equality100

sin 30◦=

141

sin C, we get

sin C =141 sin 30◦

100=

141

50= 0.705.

The equality sin C = 0.705 has two solutions:

C1 =180◦

π· sin−1(0.705) ' 44.83◦;

C2 = 180◦C1 ' 135.17◦

If we use the first solution, we get

A1 = 180◦ − B − C1 ' 180◦ − 30◦ − 44.83◦ = 105.17◦,

and usinga

sin 105.17◦=

100

sin 30◦we get

a1 =100 sin 105.17◦

sin 30◦' 193.03.

If we use the second solution, we get

A2 = 180◦ − B − C2 ' 180◦ − 30◦ − 135.17◦ = 14.83◦,

and usinga

sin 14.83◦=

100

sin 30◦we get

a2 =100 sin 14.83◦

sin 30◦' 51.19

Conclusion: This problem has two solutions:A1 ' 105.17◦

C1 ' 44.83◦

a1 ' 193.03and

A2 ' 14.83◦

C1 ' 135.17◦

a1 ' 51.19

Problem 10. Find the focus, vertex and directrix for the parabola 4(y + 1) = (x− 2)2.

Since there is no y2, the parabola has vertical symmetry axis, so its equation looks like

4p(y − k) = (x− h)2.

We have k = −1, h = 2 and 4p = 4, so we get p = 1. The vertex V (h, k) then has thecoordinates (2,−1). The focus F (h, k + p) will have coordinates (2, 0). The directrix willhave the equation y = h− p, which is

y = −2.

4