Relative to the $42^\circ$ angle, the $15$ cm side is adjacent, and you want the hypotenuse. From SOH-CAH-TOA, which trigonometric ratio connects the adjacent side and hypotenuse?

Write an equation of the form $\cos 42^\circ = \dfrac{15}{h}$, then rearrange to solve for $h$. After that, use your calculator (in degrees) and round your result to the nearest tenth.

In Desmos, type the expression 15/cos(42°) exactly like that. Make sure Desmos is set to degree mode or that you include the degree symbol (°) after 42.

Press Enter and look at the value Desmos outputs for the expression 15/cos(42°). That value (in centimeters) is the length of the hypotenuse; round it to the nearest tenth and match it to the closest answer choice.

You have a right triangle with one acute angle of $42^\circ$.

The side of length $15$ centimeters is adjacent to the $42^\circ$ angle (it touches the angle but is not the hypotenuse). You are asked to find the hypotenuse (the side opposite the right angle).

Relative to the $42^\circ$ angle, you know the adjacent side and want the hypotenuse.

From SOH-CAH-TOA:

• CAH: $\cos(\text{angle}) = \dfrac{\text{adjacent}}{\text{hypotenuse}}$.

So you should use the cosine function.

Let $h$ be the length of the hypotenuse.

Write the cosine equation using adjacent $=15$ and hypotenuse $=h$:

Solve for $h$ by multiplying both sides by $h$ and then dividing by $\cos 42^\circ$:

This expression gives the hypotenuse length in centimeters.

Use a calculator in degree mode to compute the expression $15/\cos(42^\circ)$.

The calculator value is approximately $20.2$. So the hypotenuse is about $20.2$ centimeters, which corresponds to answer choice C) 20.2.