How to find the sine of an angle - ACT Math

Card 0 of 63

Question

Solve for $\sin^2{Q}+3\sin{Q}=-2$ over the interval $0\leq Q \leq 2\pi$

Answer

Substitute x = sinQ and solve the new equation x2 + 3x = –2 by factoring. Be sure to change variables back to Q. As a result, sinQ = –1 or sinQ = –2. This function is bounded between –1 and 1 so sinQ can never be –2 and sinQ is –1 only at 3π/2 or 270 °.

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Question

See right triangle ABC. If the length AB is 8 and the length of BC is 6, what is the sine of angle A?

Answer

Sine A = Opposite / Hypotenuse = BC / AC

To find AC, use Pythagorean Theorum

AB2 + BC2 = AC2

82 + 62 = AC2

64 + 36 = AC2

100 = AC2

AC = 10

Sine A = BC / AC = 6 / 10 = 0.6

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Question

Triangle shown is a right triangle. If line and line , what is the sine of the angle at ?

Answer

$Sine(A)=\frac{Opp}{Hyp}=\frac{BC}{AC}=\frac{5}{AC}$

Now solve for using Pythagorean Theorem:

$AB^{2}+BC^{2}=AC^{2}$

$12^{2}+5^{2}=AC^{2}$

$144+25=AC^{2}$

$169=AC^{2}$

$Sine(A)=\frac{5}{13}$

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Question

If $\small a=3$ , $\small b=4$ , and $\small c=5$ , what is the sine of $\small \angle A$ ?

Answer

Recall that sin = opposite / hypotenuse. Based on the figure shown, we see that is the opposite side needed and is the hypotenuse. Plug these values in to solve.

$\small \sin \angle A = \frac{a}{c} = \frac{3}{5}$

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Question

What is the sine of $\angle ACB$ ?

Answer

Sine can be found using the SOH CAH TOA method. For sine we do $\frac{opposite}{hypotenuse}$ .

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Question

If $tan \theta = -5/12$ , and if $\theta$ is an angle between and degrees, which of the following equals $sin \theta$ ?

Answer

An angle between and degrees means that the angle is located in the second quadrant.

The tangent function is derived from taking the side opposite to the angle and dividing by the side adjacent to the angle (, as shown in the image).

Hence, the side is units long and side is units high. Therefore, according to Pythagorean Theorem rules, the side must be units long (since ).

The sine function is positive in the second quadrant. It is also equivalent to the side opposite the angle () divided by the hypotenuse ().

This makes $sin\theta = 5/13$ .

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Question

A sine function has a period of , a -intercept of , an amplitude of and no phase shift. These describe which of these equations?

Answer

Looking at this form of a sine function:

We can draw the following conclusions:

because the amplitude is specified as .

$b = \pi$ because of the specified period of since $\frac{2\pi}{\pi} = 2$ .

because the problem specifies there is no phase shift.

because the -intercept of a sine function with no phase shift is .

Bearing these in mind, $f(x) = 2sin(\pi x)$ is the only function that fits all four of those.

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Question

Solve for $\sin^2{Q}+3\sin{Q}=-2$ over the interval $0\leq Q \leq 2\pi$

Answer

Substitute x = sinQ and solve the new equation x2 + 3x = –2 by factoring. Be sure to change variables back to Q. As a result, sinQ = –1 or sinQ = –2. This function is bounded between –1 and 1 so sinQ can never be –2 and sinQ is –1 only at 3π/2 or 270 °.

Compare your answer with the correct one above

Question

See right triangle ABC. If the length AB is 8 and the length of BC is 6, what is the sine of angle A?

Answer

Sine A = Opposite / Hypotenuse = BC / AC

To find AC, use Pythagorean Theorum

AB2 + BC2 = AC2

82 + 62 = AC2

64 + 36 = AC2

100 = AC2

AC = 10

Sine A = BC / AC = 6 / 10 = 0.6

Compare your answer with the correct one above

Question

Triangle shown is a right triangle. If line and line , what is the sine of the angle at ?

Answer

$Sine(A)=\frac{Opp}{Hyp}=\frac{BC}{AC}=\frac{5}{AC}$

Now solve for using Pythagorean Theorem:

$AB^{2}+BC^{2}=AC^{2}$

$12^{2}+5^{2}=AC^{2}$

$144+25=AC^{2}$

$169=AC^{2}$

$Sine(A)=\frac{5}{13}$

Compare your answer with the correct one above

Question

If $\small a=3$ , $\small b=4$ , and $\small c=5$ , what is the sine of $\small \angle A$ ?

Answer

Recall that sin = opposite / hypotenuse. Based on the figure shown, we see that is the opposite side needed and is the hypotenuse. Plug these values in to solve.

$\small \sin \angle A = \frac{a}{c} = \frac{3}{5}$

Compare your answer with the correct one above

Question

What is the sine of $\angle ACB$ ?

Answer

Sine can be found using the SOH CAH TOA method. For sine we do $\frac{opposite}{hypotenuse}$ .

Compare your answer with the correct one above

Question

If $tan \theta = -5/12$ , and if $\theta$ is an angle between and degrees, which of the following equals $sin \theta$ ?

Answer

An angle between and degrees means that the angle is located in the second quadrant.

The tangent function is derived from taking the side opposite to the angle and dividing by the side adjacent to the angle (, as shown in the image).

Hence, the side is units long and side is units high. Therefore, according to Pythagorean Theorem rules, the side must be units long (since ).

The sine function is positive in the second quadrant. It is also equivalent to the side opposite the angle () divided by the hypotenuse ().

This makes $sin\theta = 5/13$ .

Compare your answer with the correct one above

Question

A sine function has a period of , a -intercept of , an amplitude of and no phase shift. These describe which of these equations?

Answer

Looking at this form of a sine function:

We can draw the following conclusions:

because the amplitude is specified as .

$b = \pi$ because of the specified period of since $\frac{2\pi}{\pi} = 2$ .

because the problem specifies there is no phase shift.

because the -intercept of a sine function with no phase shift is .

Bearing these in mind, $f(x) = 2sin(\pi x)$ is the only function that fits all four of those.

Compare your answer with the correct one above

Question

Solve for $\sin^2{Q}+3\sin{Q}=-2$ over the interval $0\leq Q \leq 2\pi$

Answer

Substitute x = sinQ and solve the new equation x2 + 3x = –2 by factoring. Be sure to change variables back to Q. As a result, sinQ = –1 or sinQ = –2. This function is bounded between –1 and 1 so sinQ can never be –2 and sinQ is –1 only at 3π/2 or 270 °.

Compare your answer with the correct one above

Question

See right triangle ABC. If the length AB is 8 and the length of BC is 6, what is the sine of angle A?

Answer

Sine A = Opposite / Hypotenuse = BC / AC

To find AC, use Pythagorean Theorum

AB2 + BC2 = AC2

82 + 62 = AC2

64 + 36 = AC2

100 = AC2

AC = 10

Sine A = BC / AC = 6 / 10 = 0.6

Compare your answer with the correct one above

Question

Triangle shown is a right triangle. If line and line , what is the sine of the angle at ?

Answer

$Sine(A)=\frac{Opp}{Hyp}=\frac{BC}{AC}=\frac{5}{AC}$

Now solve for using Pythagorean Theorem:

$AB^{2}+BC^{2}=AC^{2}$

$12^{2}+5^{2}=AC^{2}$

$144+25=AC^{2}$

$169=AC^{2}$

$Sine(A)=\frac{5}{13}$

Compare your answer with the correct one above

Question

If $\small a=3$ , $\small b=4$ , and $\small c=5$ , what is the sine of $\small \angle A$ ?

Answer

Recall that sin = opposite / hypotenuse. Based on the figure shown, we see that is the opposite side needed and is the hypotenuse. Plug these values in to solve.

$\small \sin \angle A = \frac{a}{c} = \frac{3}{5}$

Compare your answer with the correct one above

Question

What is the sine of $\angle ACB$ ?

Answer

Sine can be found using the SOH CAH TOA method. For sine we do $\frac{opposite}{hypotenuse}$ .

Compare your answer with the correct one above

Question

If $tan \theta = -5/12$ , and if $\theta$ is an angle between and degrees, which of the following equals $sin \theta$ ?

Answer

An angle between and degrees means that the angle is located in the second quadrant.

The tangent function is derived from taking the side opposite to the angle and dividing by the side adjacent to the angle (, as shown in the image).

Hence, the side is units long and side is units high. Therefore, according to Pythagorean Theorem rules, the side must be units long (since ).

The sine function is positive in the second quadrant. It is also equivalent to the side opposite the angle () divided by the hypotenuse ().

This makes $sin\theta = 5/13$ .

Compare your answer with the correct one above

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