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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

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8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Trigonometry

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

Possible Answers:

$(2,-\sqrt2)$

$(-\sqrt2,-\sqrt2)$

$(\sqrt2,\sqrt2)$

$(\sqrt2,-\sqrt2)$

(2,2)

Correct answer:

$(\sqrt2,-\sqrt2)$

Explanation:

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 r = 2 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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Example Question #1 : Trigonometry

Triangle

What is the $\sin C$ ?

Possible Answers:

$\frac{BC}{AB}$

$\frac{AB}{BC}$

$\frac{AB}{AC}$

$\frac{AC}{AB}$

$\frac{BC}{AC}$

Correct answer:

$\frac{AB}{AC}$

Explanation:

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

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

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Example Question #1 : Trigonometry

Triangle

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

Possible Answers:

$x\cos\theta$

$x\sin\theta$

$\cos(x\times \theta)$

$x\tan\theta$

$\frac{\sin(x)}{\theta}$

Correct answer:

$x\cos\theta$

Explanation:

$\cos\theta$ 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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Example Question #2 : Understanding Sine, Cosine, And Tangent

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

Possible Answers:

$\sqrt{3}$

$\frac{1}{3}$

$\frac{1}{2}$

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

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

Correct answer:

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

Explanation:

The pattern for the side of a 30:60:90 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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Example Question #3 : Understanding Sine, Cosine, And Tangent

If sin(x)=0.5 , what is sin(8x) if is between and $2\pi$ ?

Possible Answers:

0.5

-1.5

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

$-\frac{\sqrt{3}}{2}$

-0.5

Correct answer:

$-\frac{\sqrt{3}}{2}$

Explanation:

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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Example Question #2 : Trigonometry

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

Possible Answers:

$\frac{4}{3}$

$\frac{5}{4}$

$\frac{3}{4}$

$\frac{5}{3}$

$\frac{3}{5}$

Correct answer:

$\frac{5}{3}$

Explanation:

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:

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

O^2+4^2=5^2

O^2+16=25

O^2=9

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

O=3

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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Example Question #1 : Trigonometry

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

Possible Answers:

$\cos{x}$

$\csc{x}$

$\sec{x}$

$\cot{x}$

$\sin x$

Correct answer:

$\sec{x}$

Explanation:

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

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Example Question #1 : Arcsin, Arccos, Arctan

Trig_id

What is $\theta$ if o=10 and a=8 ?

Possible Answers:

$0.90^{\circ}$

$51.34^{\circ}$

$55.14^{\circ}$

$89.60^{\circ}$

$53.14^{\circ}$

Correct answer:

$51.34^{\circ}$

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.

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Example Question #1 : Trigonometry

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

What is ?

Possible Answers:

$60^\circ$

$30^\circ$

$26.57^\circ$

$0.009^\circ$

$1^\circ$

Correct answer:

$30^\circ$

Explanation:

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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Example Question #4 : Trigonometry

Trig_id

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

Possible Answers:

$0.50^{\circ}$

$85.92^{\circ}$

$28.95^{\circ}$

$59.28^{\circ}$

$50.53^{\circ}$

Correct answer:

$28.95^{\circ}$

Explanation:

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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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Trigonometry

Example Question #1 : Trigonometry

Example Question #1 : Trigonometry

Example Question #2 : Understanding Sine, Cosine, And Tangent

Example Question #3 : Understanding Sine, Cosine, And Tangent

Example Question #2 : Trigonometry

Example Question #1 : Trigonometry

Example Question #1 : Arcsin, Arccos, Arctan

Example Question #1 : Trigonometry

Example Question #4 : Trigonometry

All High School Math Resources

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