1 a) Explain why sin(x) = cos(90 - x) when x represents an acute angle. b) Is it ever possible that sin(x) = cos(x)? Explain your answer.

2.
In right ΔABC, mC = 90º, if sin A = m, find cos B.

Choose:
 90 - m 45 - m 90 + m m

3.
Solve for θ (angles are acute):
a) cos 60º = sin θ
Choose:
 30º 45º 60º 90º

b) sin 71º = cos θ
Choose:
 71º 35º 29º 19º

c) sin θ = cos (θ + 20)
Choose:
 20º 35º 60º 70º

d) sin (θ - 60) = cos θ
Choose:
 75º 35º 30º 15º

 4 Given right triangle ABC with right angle C, and sin A = ¼. Which of the following expressions are also equal to ¼? Select all that apply and hit "Submit". 1. cos(A) 2. cos(B) 3. cos(90º - A) 4. cos(90º - B) 5. sin(B)

 5 In attempting to solve for x in the problem at the right, students responded with a variety of equations. Which, if any, of the following equations are correct? Select all that apply and hit "Submit". 1. 2. 3. 4. 5. 6. None are true.

6.
In right ΔABC, mC = 90º, cos A = 1/5.
What is sin B ?

Choose:
 1/5 3/5 90 - 1/5 90 - 4/5

7.
In right ΔABC, mC = 90º. Simplify the following expression:

Choose:
 30 45 90 0

8.
Given that sin (x + 10)º = cos (3x + 20)º, find the number of degrees in the acute angles of the corresponding right triangle.

Choose:
 30º and 60º 45º and 45º 25º and 65º 40º and 50º

9.
In right ΔABC, mC = 90º, sin A = 3x - 0.6 and cos B = 4x - 0.9.
Find x.

Choose:
 0.3 0.4 0.6 1.5

10.
In right ΔABC, mC = 90º and mA does not equal the m∠B.
If sin A = m and cos A = k, find cos B + sin B.

Choose:
 m + k m - k k - m 90 - (m + k)