PLANE TRIGONOMETRY EXAM III SOLUTIONS

Be low you will find 10 problems, each worth 10 points. Solve the problems in the space
provided (use the back page as scratch paper). When writing a solution to a problem , show
all work. No books or notes are allowed. Sign and submit your formula sheet with
the exam.

Problem	Score	Problem	Score
1		6
2		7
3		8
4		9
5		10

Total (out of 100 points):

Problem 1. Find the exact value of arccos

Problem 2. Compute:
(a) cos(arctan 3/4 ).

Denote arctan 3/4 by t, so that tan t = 3/4 , and Since tan t is positive , it follows
that t is in the first quadrant. This gives

so we get cos t = 4/5 .

(b) tan(2 arctan 3/4 ).

Problem 3. A triangle ABC has c = 20 in, Approximate the
remaining parts. (Ex press all angles in degrees. Round to two decimal places .)

Using the Law of Sines
we have

The equality gives

and the equality gives

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

The given line has the equation x−8 = 2y, which can be written as y = 1/2 (x−8), that is:

This means that the given line has slope

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

Using the Point-Slope Equation y = m(x − x₁) + y₁, the unknown line has the equation
y = −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 of
the 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 the
parabola is:

4(x − 2) = (y − 1)².

Problem 6. Write the complex number 2−2i in trigonometric form : r(cosθ +i sinθ ) with

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

The angle θ is then determined by

It is clear that satisfies the above equalities. So the trigonometric form is

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

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

Next we find the angle using the Law of Cosine:

Finally, we find

(b) Find the area of the triangle.
We use the formula

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:
 

(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 set
y = 0 and we solve for x. We have −2x + 8 = 0, which gives 8 = 2x, thus x = 4. The
x-intercept is the point (4, 0).

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

Using the Law of Sines, we have

Using the equality we get

The equality sinC = 0.705 has two solutions:

If we use the first solution, we get

and using we get

If we use the second solution, we get

and using we get

Conclusion: This problem has two solutions:

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 the
coordinates (2,−1). The focus F(h, k + p) will have coordinates (2, 0). The directrix will
have the equation y = h − p, which is

y = −2.