In right triangles, SOHCAHTOA tells us that \(\displaystyle \sin A = \frac{\textup{opposite}}{\textup{hypotenuse}}\), and we know that \(\displaystyle \angle A = 42^{\circ}\) and hypotenuse \(\displaystyle AC = 25\). Therefore, a simple substitution and some algebra gives us our answer.

\(\displaystyle \sin 42^{\circ} = \frac{CB}{25}\)

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

\(\displaystyle 17.5= CB\) Isolate the variable term.

Thus, \(\displaystyle 17.5= CB\).

In right triangles, SOHCAHTOA tells us that \(\displaystyle \sin A = \frac{\textup{opposite}}{\textup{hypotenuse}}\), and we know that \(\displaystyle \angle A = 33^{\circ}\) and hypotenuse \(\displaystyle AC = 5400\). Therefore, a simple substitution and some algebra gives us our answer.

\(\displaystyle \sin 33^{\circ} = \frac{CB}{5400}\)

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

\(\displaystyle 2916= CB\) Isolate the variable term. 

Thus, \(\displaystyle 2916= CB\).

In right triangles, SOHCAHTOA tells us that \(\displaystyle \sin A = \frac{\textup{opposite}}{\textup{hypotenuse}}\), and we know that \(\displaystyle \angle A = 45^{\circ}\) and hypotenuse \(\displaystyle AC = 17\). Therefore, a simple substitution and some algebra gives us our answer.

\(\displaystyle \sin 45^{\circ} = \frac{CB}{17}\)

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

Thus, \(\displaystyle 12.02= CB\).

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

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

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 °.

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

\(\displaystyle \small \sin \angle A = \frac{a}{c} = \frac{3}{5}\)

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

Now solve for \(\displaystyle AC\) using Pythagorean Theorem:

\(\displaystyle AB^{2}+BC^{2}=AC^{2}\)

\(\displaystyle 12^{2}+5^{2}=AC^{2}\)

\(\displaystyle 144+25=AC^{2}\)

\(\displaystyle 169=AC^{2}\)

\(\displaystyle AC=13\)

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

An angle between \(\displaystyle 90\) and \(\displaystyle 180\) 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 (\(\displaystyle y/x\), as shown in the image).

Hence, the side \(\displaystyle x\) is \(\displaystyle 12\) units long and side \(\displaystyle y\) is \(\displaystyle 5\) units high. Therefore, according to Pythagorean Theorem rules, the side \(\displaystyle r\) must be \(\displaystyle 13\) units long (since \(\displaystyle 5^2 + 12^2 = 13^2\)).

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

This makes \(\displaystyle sin\theta = 5/13\).

Looking at this form of a sine function:

\(\displaystyle f(x) = asin(bx + c) + d\)

We can draw the following conclusions:

• \(\displaystyle a = 2\) because the amplitude is specified as \(\displaystyle 2\).
• \(\displaystyle b = \pi\) because of the specified period of \(\displaystyle 2\) since \(\displaystyle \frac{2\pi}{\pi} = 2\).
• \(\displaystyle c = 0\) because the problem specifies there is no phase shift.
• \(\displaystyle d = 0\) because the \(\displaystyle y\)-intercept of a sine function with no phase shift is \(\displaystyle 0\).

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