1.      D     

Since the triangle is a right triangle and the figure gives the value of the angle opposite side AC, you can use the sine function to find the length of AC.

2.      B     

Since the figure gives only the values for the two legs of this right triangle, to find the measure of you’ll first need to calculate a value for tan B and then take the inverse.

which means that

3.      C     

The kite-flying situation can be modeled by a right triangle with an acute angle of 55º, and a leg opposite that angle whose length is 100 feet. Once you picture the situation as a right triangle, you can see that

where the hypotenuse is the length of the kite string. Therefore, letting x represent the hypotenuse:

4.      B     

Simplify the left side using trigonometric identities to make this problem easier to solve. First, you need to rearrange the identity sin² x + cos² x = 1 so that you find cos² x – 1 = –sin² x. Then substitute this into the equation. Later, substitute tan x for ^{sin x}/_{cos x}.

Now solve the equation tan x = 1. It’s just a matter of taking the inverse of both sides of the equation: = arctan 1 = 45º.

5.      B     

If = ¹/₄ , we can rewrite the given conditions: sin and cos 4 are greater than zero. These are the conditions we must meet.

In order for sin a to be greater than zero, 0º < < 180º, because sine is only positive in the first two quadrants. This mean 0 < 4 < 720º. Cosine, however, is positive only in the first and fourth quadrants. Thus, for 0º < x < 720º, cosine is only positive in the following intervals: (0º, 90º), (270º, 450º), and (630º, 720º). By dividing these intervals by four, the range of is defined: 0º < < 22.5º, 67.5º < < 112.5º, or 157.5º < < 180º. The only answer choice that does not fall within one of these intervals is 65º.

6.      C     

The variable b adjusts the period of the standard function from 2π to ^2π /_b. The standard sine function, y = sin x, crosses the x-axis three times in the interval 0 ≤ x ≤ 2π: at 0, π, and 2π. So, since the period of a function is the interval between each repeat of the function’s curve, the only way for the graph of y = 3 sin bx to cross the x-axis more often than y = sin x is to have a shorter period. Thus, b must be greater than 1.

To make the function cross the x-axis ⁷/₃ as many times as y = sin x does, you might be tempted to make the period ³ /₇ as long as 2π, which would correspond to a b value of ⁷/₃. This, however, is wrong, because it counts the intersection at 2π too many times. You can check this result if you have a graphing calculator.

By halving the period of the function, you might think that the number of crossings would double. Actually, y = sin 2x crosses the x-axis only five times because the crossing at 2π does not figure into the doubling.

Using this logic as a guide, you see that to achieve seven crossings, you must make the period 3 times shorter so that the first 2 crossings are tripled in number and the crossing at 2π is added at the end. This means that the period of the unknown function is ^2π/₃, and b = 3.

7.      A     

In the cosine function, the amplitude is the coefficient in front of cosine and the period is 2π divided by the coefficient of x. So for the function y = 2 cos (4x + 2) – 7, the amplitude is 2 and the period is ^2π⁄₄ = ^π⁄₂.

8.      B     

This problem takes a few steps. Your goal is to find AB and the height to vertex C. Then you can use the area formula, A = ¹⁄₂ bh, where b is the base and h is the height.

First, draw an altitude from C to AB.

The length of this altitude is the height of the triangle. In the triangle you just formed, triangle ACD, sin 40º = ^h⁄₄. So, h = 4 sin 40º ≈ 2.57. The Pythagorean theorem can now be used to find lengths AD and BD:

The sum of AD and BD is AB, approximately 9.58. Finally, you can plug these values back into the area formula: