Trigonometry

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Questions 1 - 10

In the figure below, $\overline{BD}$ is a diagonal of quadrilateral . $\overline{BD}$ has a length of . $\angle CBD$ is congruent to $\angle BDC$ .

Screen shot 2020 08 27 at 4.39.20 pm

Which of the following is a true statement?

The area of quadrilateral is $\frac{1}{2}$ .

The area of quadrilateral is .

The perimeter of quadrilateral is .

Explanation

Since and are perpendicular, $\angle BCD$ is a right angle. Since no triangle can have more than one right angle, and $\Delta BCD$ is isosceles, $\angle CBD$ must be congruent to $\angle BDC$ . Since angle CBD is congruent to $\angle BDC$ and $\angle BCD$ measures 90 degrees, $\angle CBD$ and $\angle BDC$ can be calculated as follows:

Therefore, and are both equal to 45 degrees. $\Delta BCD$ is a 45-45-90 triangle. Therefore, the ratio between side lengths and hypotenuse is $1:\sqrt{2}$ . Anyone of the four side lengths of quadrilateral must, therefore, be equal to $\frac{BD}{\sqrt{2}}=\frac{1}{\sqrt{2}}$ . To find the area of , multiply two side lengths: $\frac{1}{\sqrt{2}}*\frac{1}{\sqrt{2}}=\frac{1}{2}$ .

Suppose the distance from a student's eyes to the floor is 4 feet. He stares up at the top of a tree that is 20 feet away, creating a 30 degree angle of elevation. How tall is the tree?

$\frac{20\sqrt{3}+12}{3}$

$\frac{20\sqrt{3}+4}{3}$

$\frac{20\sqrt{3}+12}{3}$

$10\sqrt3+4$

Explanation

The height of the tree requires using trigonometry to solve. The distance of the student to the tree , partial height of the tree , and the distance between the student's eyes to the top of the tree will form the right triangle.

The tangent operation will be best used for this scenario, since we have the known distance of the student to the tree, and the partial height of the tree.

Set up an equation to solve for the partial height of the tree.

$tan(\theta) = \frac{\textup{Opposite side to angle}}{\textup{Adjacent side to angle}}$

$tan(30) = \frac{P}{20}$

Multiply by 20 on both sides.

$P=20tan(30) = 20(\frac{\sqrt{3}}{3}) = \frac{20\sqrt{3}}{3}$

We will need to add this with the height of the student's eyes to the ground to get the height of the tree.

$\frac{20\sqrt{3}}{3} + 4 = \frac{20\sqrt{3}}{3} + \frac{12}{3}$

The answer is: $\frac{20\sqrt{3}+12}{3}$

What angle is complementary to $34^\circ$ ?

$56^\circ$

$66^\circ$

$34^\circ$

$146^\circ$

$156^\circ$

Explanation

Two complementary angles add up to $90^\circ$ .

Therefore, $34^\circ+x^\circ=90^\circ$ .

$x^\circ=90^\circ-34^\circ$

$x^\circ=56^\circ$

How many radians are in $270^\circ$ ?

$\frac{3\pi}{2}$

$2\pi$

$\pi$

$\frac{2\pi}{3}$

$\frac{5\pi}{2}$

Explanation

The conversion for radians is $180^\circ=\pi$ , so we can make a ratio:

$\frac{270^\circ}{x}=\frac{180^\circ}{\pi}$

Cross multiply:

$270^\circ*\pi=180^\circ*x$

Isolate :

$\frac{270^\circ}{180^\circ}\pi=x$

$\frac{3\pi}{2}=x$

If Angle equals $72^{\circ}$ , what is the equivalent angle in radians (to the nearest hundredth)?

$1.26\ radians$

$4125.29\ radians$

$4.125\ radians$

$1256.6\ radians$

$0.628\ radians$

Explanation

To convert between radians and degrees, it is important to remember that:

$\pi, rad = 180^{\circ}$

With this relationship in mind, we can convert from degrees to radians with the following formula:

$72^{\circ}\cdot \frac{\pi}{180^{\circ}} = 1.26,rad$

Simplify

$\frac{\sin^{2}{\left (x \right )} + \cos^{3}{\left (x \right )}}{2 \cos^{2}{\left (x \right )} - 1}$

$\frac{1}{2} \left(\cos^{3}{\left (x \right )} \csc^{2}{\left (x \right )} + 1\right)$

$\frac{1}{2} \cos^{3}{\left (x \right )} \csc^{2}{\left (x \right )}$

$\cos^{2}{\left (x \right )} + \csc^{3}{\left (x \right )}$

$\sin^{2}{\left (x \right )} + \sec^{2}{\left (x \right )}$

$\frac{1}{2}$

Explanation

The first step to simplifying is to remember an important trig identity.

$\sin^{2}{\left (x \right )} + \cos^{2}{\left (x \right )} = 1$

If we rewrite it to look like the denominator, it is.

$\sin^{2}{\left (x \right )} = \cos^{2}{\left (x \right )} - 1$

Now we can substitute this in the denominator.

$\frac{\sin^{2}{\left (x \right )} + \cos^{3}{\left (x \right )}}{2 \sin^{2}{\left (x \right )}}$

Now write each term separately.

$\frac{1}{2} + \frac{\cos^{3}{\left (x \right )}}{2 \sin^{2}{\left (x \right )}}$

Remember the following identities.

$\\csc{\left (x \right )} = \frac{1}{\sin{\left (x \right )}} \\\sec{\left (x \right )} = \frac{1}{\cos{\left (x \right )}} \\\tan{\left (x \right )} = \frac{\sin{\left (x \right )}}{\cos{\left (x \right )}} \\\cot{\left (x \right )} = \frac{\cos{\left (x \right )}}{\sin{\left (x \right )}}$

Now simplify, and combine each term.

$\\frac{1}{2} \cos^{3}{\left (x \right )} \csc^{2}{\left (x \right )} + \frac{1}{2} \\\frac{1}{2} \left(\cos^{3}{\left (x \right )} \csc^{2}{\left (x \right )} + 1\right)$

Trig_id

What is $\theta$ if and ?

$51.34^{\circ}$

$53.14^{\circ}$

$55.14^{\circ}$

$0.90^{\circ}$

$89.60^{\circ}$

Explanation

In order to find $\theta$ we need to utilize the given information in the problem. We are given the opposite and adjacent sides. We can then, by definition, find the $\tan$ of $\theta$ and its measure in degrees by utilizing the $\arctan$ function.

$\tan=\frac{opposite}{adjacent}$

$\tan=\frac{10}{8}$

$\tan=1.25$

Now to find the measure of the angle using the $\arctan$ function.

$\theta=\arctan1.25$

$\rightarrow 51.34^{\circ}$

If you calculated the angle's measure to be $0.90^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

Using trigonometric identities prove whether the following is valid:

$\sin^{2}A + \cot^{2}A = \frac{2\sin^{4}A +2\cos^{2}A}{1 - \cos 2A}$

True

False

Uncertain

Only in the range of: $0 < A < \frac{\pi}{2}$

Only in the range of: $0 < A < 2\pi$

Explanation

We can work with either side of the equation as we choose. We work with the right hand side of the equation since there is an obvious double angle here. We can factor the numerator to receive the following:

$\sin^{2}A + \cot^{2}A = \frac{2\left(\sin^{4}A + \cos^{2}A\right)}{1-\cos 2A}$

Next we note the power reducing formula for sine so we can extract the necessary components as follows:

$\sin^{2}A + \cot^{2}A = \frac{2}{1-\cos 2A}\left(\sin^{4}A + \cos^{2}A\right)$

The power reducing formula must be inverted giving:

$\sin^{2}A + \cot^{2}A = \frac{1}{\sin^{2}A}\left(\sin^{4}A + \cos^{2}A\right)$

Now we can distribute and reduce:

$\sin^{2}A + \cot^{2}A = \sin^{2}A + \frac{\cos^{2}A}{\sin^{2}A}$

Finally recalling the basic identity for the cotangent:

$\sin^{2}A + \cot^{2}A = \sin^{2}A + \cot^{2}A$

This proves the equivalence.

Which of the following is the correct definition of a phase shift?

A measure of the length of a function between vertical asymptotes

The distance a function is shifted diagonally from the general position

The distance a function is shifted horizontally from the general position

The distance a function is shifted vertically from the general position

Explanation

Take the function for example. The graph for is

If we were to change the function to $cos(x + \pi)$ , our phase shift is $-\pi$ . This means we need to shift our entire graph $\pi$ units to the left.