Trigonometric Operations - Math

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What is $\theta$ if and ?

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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 then your calculator was set to radians and needs to be set on degrees.

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

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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.

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

What is ?

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

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

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

$x^\circ=30^\circ$

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

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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:

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}$$

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

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

What is the $\sin C$ ?

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

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

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

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$\cos\theta$ 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.

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

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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 $(\sqrt2,-\sqrt2)$ .

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

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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}$

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

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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}$}{2}$ .

What is $\theta$ if and ?

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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 then your calculator was set to radians and needs to be set on degrees.

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

Tap to see back →

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.

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

What is ?

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

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

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

$x^\circ=30^\circ$

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

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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:

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}$$

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

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

What is the $\sin C$ ?

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

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

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

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$\cos\theta$ 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.

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

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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 $(\sqrt2,-\sqrt2)$ .

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

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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}$

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

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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}$}{2}$ .