Math 102, Trigonometry PALS
Test # 1 Spring 2016
Professor Marian Smith-Subbarao
_______ __________________
Name
Score: _______________
Percent: ______________
Directions:
 Show all work and clearly mark your answers.

Partial credit may be given, even if the answer is incorrect, if your work is clear – attach
additional scratch pages you wish to be considered. If you work on scratch paper, please put
the problem number by your work

If you do not show your work, you may not get credit.

Unless otherwise instructed, leave all answers as fractions; improper fractions are OK.

Simplify radicals, but do not compute them.
NO: Telephones, Books, Notes; CALCULATORS ARE NOT ALLOWEED
Suggestions:
 Choose the problems you understand best to work first.
 If you get stuck, write down what you do understand for partial credit and move on
 Show your work clearly
 Check your solutions
 Evaluate your solutions for “reasonableness”
All Triangles refer to the following sketch: angles A, B, C, sides a, b, c. Assume a right or
isosceles triangle only if it is specified.
C
b
a
B
A
c
Find the 𝒙- and 𝒚-intercepts of the given equation. (10 points)
1. 𝑦 = 𝑥 3 − 𝑥 2 − 12𝑥
x intercept, y = 0: 0 = 𝑥 3 − 𝑥 2 − 12𝑥 = x(x2 – x – 12) = x(x-4)(x+3); x = 0, -3, 4
y intercept, x = 0; y = 0
Find all real solutions of the given equations. (8 points)
2. 4|𝑥 − 9| + 7 = 2
4|x-9| = -5; no real solutions
3. 𝑥 4 − 7𝑥 2 = −12
Let y = x2, y2 – 7 + 12 = 0 = (y-3)(y-4); y = 3, y = 4, x = ± 2, ±√3
4. 8𝑥 −2 − 2𝑥 −1 − 3 = 0
Let y = x-1 , we have 8y2 -2y – 3 = 0 = (2y + 1)(4y - 3), y = -1/2, y = 3/4 or x = -2, 4/3
5.
5𝑥−3
2
< 4 – 2x <
8𝑥−3
3
5x-3 < 8 – 4x, 9x<11, x<11/9;
15/14 < x < 11/9
6.
|
3−5𝑥
4
12 – 6x < 8x-3, 14x>15, x>15/14
|<2
|3-5x|<8 so 3-5x < 8 or 5x>-5, x>-1 and 3-5x>-8, -5x>-11, x<11/5
-1<x<11/5
7. If sin A = -1, find the remaining trig functions. (6 points)
270 deg. Cos A = 0, tan A undef, csc A = -1, sec A undef, cot = 0
8. In triangle ABC, where C is a right angle, angle A = 3x - 4 and angle B = 2x - 1. Find
angles A and B. (6 points)
Have A + B = 90, 3x – 4 + 2x – 1 = 90 = 5x-5, 5x=95, x = 19; A = 53, B = 37
9. If sin A = 2/5, and tan A < 0, find the value of the other five trig functions of A.
(Remember: fractions and radicals are OK in your answer. However, please express your
answer in the simplest form possible.) (10 points)
because of the signs, are in the 2nd quadrant
opp/hyp = 2/5, if let a = adj, a2 = 25 – 4 = 21, a = √21
cos = −√21/5, sec = -5/√21, csc = 5/2, tan = -2/√21, cot = - √21/2
10. If angle B is in standard position and cos B = 3/5, find the equation of the line that forms
the terminal side of angle B. (8 points)
Need the other side of the triangle; 25 – 9 = 16, side is 4
y value is 4, x is 3, so tan B = 4/3
y = 4/3 x
11. Find cot θ if csc θ = sqrt(37) / 6 and θ is in quadrant I. (10 points)
csc is h/opp, h2=37 and opp2 = 36. So we have: 37 = 36 + adj2, adj = 1
Cot = adj/opp = 1/6
12. Given arbitrary triangles ABC and DEF, angle A = angle D, and angle B = angle E, if a=
3, c = 5, d = 12, find side f. (10 points)
Two angles are equal, so we have similar triangles
a/d = b/e = c/f. Since we have a/d = 3/12, we must have c/f = 5/f = 3/12; 3f = 60, f = 20
13. Find the exact value of x in the figure below. (15 points)
The left hand triangle is 30-60-90, and the right 45-45-90
We have the length 40, which is the hypotenuse of the 30 deg angle. X is the hypotenuse
of the 45 degree angle.
The vertical line is unknown length, but is equal for the two triangles; let it be y
cos 30 = y/40 = √3 /2, so y = 20 √3 .
y/x = sin 45 = 1/√2 ; since y/x =1/√2 x=√2 y = 20 √6