How to find the sine of an angle

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ACT Math › How to find the sine of an angle

Questions 1 - 7

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

$f(x) = 2sin(\pi x)$

$f(x) = sin(\pi x)$

$f(x) = 2sin(2\pi x)$

$f(x) = 2sin(\pi x) + 1$

Explanation

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.

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

$\small \frac{3}{5}$

$\small \frac{4}{5}$

$\small \frac{5}{3}$

$\small \frac{4}{3}$

$\small \frac{3}{4}$

Explanation

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.

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

Triangle

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

0.6

0.8

Explanation

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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What is the sine of $\angle ACB$ ?

$\frac{5}{13}$

$\frac{5}{12}$

$\frac{12}{13}$

$60^{\circ}$

$\frac{12}{5}$

Explanation

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

Triangle

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

$\frac{5}{13}$

$\frac{5}{12}$

$\frac{12}{5}$

$\frac{12}{13}$

$\frac{13}{5}$

Explanation

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

Now solve for using Pythagorean Theorem:

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

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

$144+25=AC^{2}$

$169=AC^{2}$

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

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

Q = 3π or does not exist 2

Q = π or 3π 2 2

Q = π or does not exist 2

Q = π or 2π

Explanation

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 °.

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