Question

right triangle abc is shown.
which equation can be used to solve for c?
image of right triangle abc with right angle at c, ac = 3 m, hypotenuse c, angle at b is 50 degrees, side bc is labeled a
options:

• $sin(50^circ) = \frac{3}{c}$
• $sin(50^circ) = \frac{c}{3}$
• $cos(50^circ) = \frac{c}{3}$
• $cos(50^circ) = \frac{3}{c}$

Explanation:

Step1: Recall SOHCAHTOA

In a right triangle, $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$.

Step2: Identify sides relative to \(50^\circ\)

In \(\triangle ABC\), \(\angle B = 50^\circ\), right - angled at \(C\). The side opposite to \(\angle B\) is \(AC = 3\) m, and the hypotenuse is \(AB=c\).
Using the sine formula \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\), for \(\theta = 50^\circ\), we have \(\sin(50^{\circ})=\frac{AC}{AB}=\frac{3}{c}\).
We can check the other options:

• For the option \(\sin(50^{\circ})=\frac{c}{3}\), since \(c\) is the hypotenuse and \(3\) is the opposite side, \(\sin(50^{\circ})\) should be \(\frac{3}{c}\) not \(\frac{c}{3}\), so this is wrong.
• For the option \(\cos(50^{\circ})=\frac{c}{3}\), \(\cos(50^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), the adjacent side to \(\angle B\) is \(BC\), not \(c\) (hypotenuse), so this is wrong.
• For the option \(\cos(50^{\circ})=\frac{3}{c}\), \(\cos(50^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), the adjacent side to \(\angle B\) is \(BC\), not \(3\) (opposite side), so this is wrong.

Answer:

\(\boldsymbol{\sin(50^{\circ})=\frac{3}{c}}\) (the first option)