Trigonometric Operations - High School Math

Card 0 of 10

Question

What is $\theta$ if and ?

Answer

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.

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Question

$o=7.6, h=15.7, \Theta = ?$

Answer

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

$\sin=\frac{opposite}{hypotenuse}$

$\sin=\frac{7.6}{15.7}$

$\sin=0.4841$

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

$\theta=\arcsin0.4841$

$\rightarrow 28.95^{\circ}$

If you calculated the angle's measure to be $\dpi{100} 0.50^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

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Question

$\sin(x^\circ)=\frac{1}{2}$

What is ?

Answer

To get rid of $\sin(x^\circ)$ , we take the $\arcsin$ or $\sin^{-1}$ of both sides.

$\sin(x^\circ)=\frac{1}{2}$

$\sin^{-1}(\sin(x^\circ))=\sin^{-1}(\frac{1}{2})$

$x^\circ=\sin^{-1}(\frac{1}{2})$

$x^\circ=30^\circ$

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Question

An angle has a cosine of $\frac{4}{5}$ . What will its cosecant be?

Answer

The problem tells us that the cosine of the angle will be $\frac{4}{5}$ . Cosine is the adjacent over the hypotenuse. From here we can use the Pythaogrean theorem:

$\text{Opposite}^2+\text{Adjacent}^2=\text{Hypotenuse}^2$

$\sqrt{O^2}=\sqrt{9}$

Now we know our opposite, adjacent, and hypotenuse.

The cosecant is $\csc=\frac{\text{Hypotenuse}}{\text{Opposite}}$ .

From here we can plug in our given values.

$\csc=\frac{5}{3}$

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Question

Which of these is equal to $\frac{\text{hypotenuse}}{\text{adjacent}}$ for angle ?

Answer

$\sec{x}=\frac{\text{hypotenuse}}{\text{adjacent}}$ , as it is the inverse of the $\cos{x}$ function. This is therefore the answer.

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Question

What is the $\sin C$ ?

Answer

$\sin X = \frac{OPPOSITE}{HYPOTENUSE}$

$\sin C = \frac{AB}{AC}$

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Question

In the right triangle above, which of the following expressions gives the length of y?

Answer

$\cos\theta$ is defined as the ratio of the adjacent side to the hypotenuse, or in this case $\frac{y}{x}$ . Solving for y gives the correct expression.

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Question

If the polar coordinates of a point are $\left (2,\frac{-\pi}{4} \right )$ , then what are its rectangular coordinates?

Answer

The polar coordinates of a point are given as $(r,\theta )$ , where r represents the distance from the point to the origin, and $\theta$ represents the angle of rotation. (A negative angle of rotation denotes a clockwise rotation, while a positive angle denotes a counterclockwise rotation.)

The following formulas are used for conversion from polar coordinates to rectangular (x, y) coordinates.

$x=r\cos\theta$

$y=r\sin\theta$

In this problem, the polar coordinates of the point are $\left (2,\frac{-\pi}{4} \right )$ , which means that and $\theta=-\frac{\pi}{4}$ . We can apply the conversion formulas to find the values of x and y.

$x=r\cos\theta=2\cos\left (\frac{-\pi}{4} \right )=2\cdot \frac{\sqrt2}{2}= \sqrt{2}$

$y=r\sin\theta=2\sin\left (\frac{-\pi}{4} \right )=2\cdot \frac{-\sqrt2}{2}= -\sqrt{2}$

The rectangular coordinates are $(\sqrt2,-\sqrt2)$ .

The answer is $(\sqrt2,-\sqrt2)$ .

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Question

What is the cosine of $30^\circ$ ?

Answer

The pattern for the side of a triangle is $x:x\sqrt{3}:2x$ .

Since $\cos=\frac{\text{adjacent}}{\text{hypotenuse}}$ , we can plug in our given values.

$\cos=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\cos=\frac{x\sqrt{3}}{2x}$

Notice that the 's cancel out.

$\cos=\frac{\sqrt{3}}{2}$

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Question

If , what is if is between and $2\pi$ ?

Answer

Recall that $sin(\frac{\pi}{6}) =0.5$ .

Therefore, we are looking for $sin(8*\frac{\pi}{6})$ or $sin(\frac{4\pi}{3})$ .

Now, this has a reference angle of $\frac{\pi}{3}$ , but it is in the third quadrant. This means that the value will be negative. The value of $sin(\frac{\pi}{3})$ is $\frac{\sqrt{3}}{2}$ . However, given the quadrant of our angle, it will be $-\frac{\sqrt{3}}{2}$ .

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