
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 kiteflying 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 sin^{2} x +
cos^{2} x = 1 so that you
find cos^{2 }x – 1 = –sin^{2 }x.
Then substitute this into the equation. Later, substitute tan x for
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 =
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
To make the function cross the xaxis
By halving the period of the function, you might think that the number of crossings would double. Actually, y = sin 2x crosses the xaxis 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
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:
