Sine - ACT Math

Card 0 of 405

Question

If , what is ? Round to the nearest hundredth.

Answer

Recall that the sine wave is symmetrical with respect to the origin. Therefore, for any value , the value for is . Therefore, if is , then for , it will be .

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Question

In a right triangle, cos(A) = $\frac{11}{14}$ . What is sin(A)?

Answer

In a right triangle, for sides a and b, with c being the hypotenuse, $a^{2} + b ^{2} = c^{2}$ . Thus if cos(A) is $\frac{11}{14}$ , then c = 14, and the side adjacent to A is 11. Therefore, the side opposite of angle A is the square root of $14^{2} - 11^{2}$ , which is $\sqrt{75} = 5\sqrt{3}.$ Since sin is $\frac{opposite}{hypotenuse}$ , sin(A) is $\frac{5\sqrt{3}}{14}$ .

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Question

A function with period P will repeat on intervals of length P, and these intervals are referred to as periods.

Find the period of

$sin, (\pi x)$ .

Answer

For the function

the period is equal to

$\frac{2\pi}{A}$

in this case

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

which reduces to .

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Question

You have a 30-60-90 triangle. If the hypotenuse length is 8, what is the length of the side opposite the 30 degree angle?

Answer

sin(30º) = ½

sine = opposite / hypotenuse

½ = opposite / 8

Opposite = 8 * ½ = 4

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Question

If a right triangle has a 30 degree angle, and the opposite leg of the 30 degree angle has a measure of 12, what is the value of the hypotenuse?

Answer

Use SOHCAHTOA. Sin(30) = 12/x, then 12/sin(30) = x = 24.

You can also determine the side with a measure of 12 is the smallest side in a 30:60:90 triangle. The hypotenuse would be twice the length of the smallest leg.

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Question

The radius of the above circle is . is the center of the circle. $\angle A = 110^{\circ}$ . Find the length of chord $\overline{BC}$ .

Answer

We can solve for the length of the chord by drawing a line the bisects the angle and the chord, shown below as $\overline{AD}$ .

In this circle, we can see the triangle $\bigtriangleup ADC$ has a hypotenuse equal to the radius of the circle ( $\overline{AC}$ ), an angle $\theta$ equal to half the angle made by the chord, and a side $\overline{CD}$ that is half the length of the chord. By using the sine function, we can solve for $\overline{CD}$ .

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

$\sin \left ( \theta\right ) = \frac{\overline{CD}}{\overline{AC}}$

$\sin \left ( 55^{\circ}\right ) = \frac{\overline{CD}}{5}$

$5\sin \left ( 55^{\circ}\right ) = \overline{CD}$

$4.1 = \overline{CD}$

The length of the entire chord is twice the length of $\overline{CD}$ , so the entire chord length is .

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Question

The above circle has a radius of and a center at . $\angle A = 127^{\circ}$ . Find the length of chord $\overline{BC}$ .

Answer

We can solve for the length of the chord by drawing a line the bisects the angle and the chord, shown below as $\overline{AD}$ .

In this circle, we can see the triangle $\bigtriangleup ADC$ has a hypotenuse equal to the radius of the circle ( $\overline{AC}$ ), an angle $\theta$ equal to half the angle made by the chord, and a side $\overline{CD}$ that is half the length of the chord. By using the sine function, we can solve for $\overline{CD}$ .

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

$\sin \left (\theta\right ) = \frac{\overline{CD}}{\overline{AC}}$

$\sin \left ( 63.5^{\circ}\right ) = \frac{\overline{CD}}{8}$

$8\sin \left ( 63.5^{\circ}\right ) = \overline{CD}$

$7.16 = \overline{CD}$

The length of the entire chord is twice the length of $\overline{CD}$ , so the entire chord length is .

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Question

A man has set up a ground-level sensor to look from the ground to the top of a $30\textup{ foot}$ tall building. The sensor must have an angle of $25 . 5$^{\circ}$$ upward to the top of the building. How far is the sensor from the top of the building? Round to the nearest inch.

Answer

Begin by drawing out this scenario using a little right triangle:

Note importantly: We are looking for $\small x$ as the the distance to the top of the building. We know that the sine of an angle is equal to the ratio of the side opposite to that angle to the hypotenuse of the triangle. Thus, for our triangle, we know:

$\small sin(25.5) =\frac{30}{x}$

Using your calculator, solve for :

$\small \small x=\frac{30}{sin(25.5)}$

This is $\small 69.6846149202954$ . Now, take the decimal portion in order to find the number of inches involved.

$\small 0.6846149202954 * 12 = 8.2153790435448$

Thus, rounded, your answer is $\small 69$ feet and $\small 8$ inches.

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Question

What is in the right triangle above? Round to the nearest hundredth.

Answer

Recall that the sine of an angle is the ratio of the opposite side to the hypotenuse of that triangle. Thus, for this triangle, we can say:

$\small sin(27.45)=\frac{x}{47.25}$

Solving for $\small x$ , we get:

$\small 47.25*sin(27.45)=x$

$\small x=21.7810391968006$ or $\small 21.78$

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Question

Below is right triangle with sides . What is $\sin(A)$ ?

Answer

To find the sine of an angle, remember the mnemonic SOH-CAH-TOA.
This means that

$\textup{sin} = \frac{\textup{opposite}}{\textup{hypotenuse}}$
$\cos = \frac{\textup{adjacent}}{\textup{hypotenuse}}$

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

We are asked to find the $\sin(A)$ . So at point we see that side is opposite, and the hypotenuse never changes, so it is always . Thus we see that
$\sin(A) = \frac{a}{c}$

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Question

In a given right triangle $\Delta ABC$ , hypotenuse and $\angle A = 42^{\circ}$ . Using the definition of $\sin$ , find the length of leg . Round all calculations to the nearest tenth.

Answer

In right triangles, SOHCAHTOA tells us that $\sin A = \frac{\textup{opposite}}{\textup{hypotenuse}}$ , and we know that $\angle A = 42^{\circ}$ and hypotenuse . Therefore, a simple substitution and some algebra gives us our answer.

$\sin 42^{\circ} = \frac{CB}{25}$

$.7 = \frac{CB}{25}$ Use a calculator or reference to approximate cosine.

Isolate the variable term.

Thus, .

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Question

In a given right triangle $\Delta ABC$ , hypotenuse and $\angle A = 45^{\circ}$ . Using the definition of $\sin$ , find the length of leg . Round all calculations to the nearest hundredth.

Answer

In right triangles, SOHCAHTOA tells us that $\sin A = \frac{\textup{opposite}}{\textup{hypotenuse}}$ , and we know that $\angle A = 45^{\circ}$ and hypotenuse . Therefore, a simple substitution and some algebra gives us our answer.

$\sin 45^{\circ} = \frac{CB}{17}$

$17\cdot \sin 45^{\circ}= CB$ Isolate the variable term.

Thus, .

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Question

In a given right triangle $\Delta ABC$ , hypotenuse and $\angle A = 33^{\circ}$ . Using the definition of $\sin$ , find the length of leg . Round all calculations to the nearest tenth.

Answer

In right triangles, SOHCAHTOA tells us that $\sin A = \frac{\textup{opposite}}{\textup{hypotenuse}}$ , and we know that $\angle A = 33^{\circ}$ and hypotenuse . Therefore, a simple substitution and some algebra gives us our answer.

$\sin 33^{\circ} = \frac{CB}{5400}$

$.54 = \frac{CB}{5400}$ Use a calculator or reference to approximate cosine.

Isolate the variable term.

Thus, .

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Question

Simplify: (sinΘ + cosΘ)2

Answer

Using the foil method, multiply. Simplify using the Pythagorean identity sin2Θ + cos2Θ = 1 and the double angle identity sin2Θ = 2sinΘcosΘ.

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Question

For the triangle , find $\angle A$ in degrees to the nearest integer

Note: The triangle is not necessarily to scale

Answer

To solve this equation, it is best to remember the mnemonic SOHCAHTOA which translates to Sin = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent. Looking at the problem statement, we are given the side opposite of the angle we are trying to find as well as the hypotenuse. Therefore, we will be using the SOH part of our mnemonic. Inserting our values, this becomes $SIN(\angle A) = \frac{8}{13}$ . Then, we can write $\angle A = SIN^{-1}\left ( \frac{8}{13} \right )$ . Solving this, we get $\angle A = 38^{\circ}$

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Question

A fifteen foot ladder is leaned up against a twelve foot building, reaching the top of the building. What is the angle made between the ladder and the ground? Round to the nearest hundredth of a degree.

Answer

You can draw your scenario using the following right triangle:

Recall that the sine of an angle is equal to the ratio of the oppositeside to the hypotenuse of the triangle. You can solve for the angle by using an inverse sine function:

$\small \small \Theta = sin^-^1(\frac{12}{15})=53.13010235415598$ or $53.13$^{\circ}$$ .

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Question

What is the value of $\small \Theta$ in the right triangle above? Round to the nearest hundredth of a degree.

Answer

Recall that the sine of an angle is equal to the ratio of the opposite side to the hypotenuse of the triangle. You can solve for the angle by using an inverse sine function:

$\Theta = sin^-^1(\frac{30}{75})=23.57817847820183$ or $23.58$^{\circ}$$ .

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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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Question

If $sin(2x)=\frac{-\sqrt{2}}{2}$ , what is the value of if $\frac{\pi}{2} \le 2x \le \frac{3\pi}{2}$ ?

Answer

Recall that the triangle appears as follows in radians:

Now, the sine of $\frac{\pi}{4}$ is $\frac{1}{\sqrt{2}}$ . However, if you rationalize the denominator, you get:

$\frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

Since $sin(2x)=\frac{-\sqrt{2}}{2}$ , we know that must be represent an angle in the third quadrant (where the sine function is negative). Adding its reference angle to $\pi$ , we get:

$\pi + \frac{\pi}{4}=\frac{5\pi}{4}$

Therefore, we know that:

$2x = \frac{5\pi}{4}$ , meaning that $x=\frac{5\pi}{8}$

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Question

What is the sine of the angle formed between the origin and the point if that angle is formed with one side of the angle beginning on the -axis and then rotating counter-clockwise to ?

Answer

You can begin by imagining a little triangle in the fourth quadrant to find your reference angle. It would look like this:

Now, to find the sine of that angle, you will need to find the hypotenuse of the triangle. Using the Pythagorean Theorem, , where and are leg lengths and is the length of the hypotenuse, the hypotenuse is $\sqrt{a^2+b^2}$ , or, for our data:

$\sqrt{4^2+10^2}=\sqrt{16+100}=\sqrt{116}=2\sqrt{29}$

The sine of an angle is:

$\frac{opposite}{hypotenuse}$

For our data, this is:

$\frac{10}{2\sqrt{29}}=\frac{5}{\sqrt{29}}$

Since this is in the fourth quadrant, it is negative, because sine is negative in that quadrant.

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