Question

right triangle abc is shown. which equation can be used to solve for c? a 3 m c a b 50° sin(50°) = (3)/(c) sin(50°) = (c)/(3) cos(50°) = (c)/(3) cos(50°) = (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 $50^\circ$

• Hypotenuse: $c$
• Opposite to $50^\circ$: $AC = 3\,\text{m}$ (since $\angle B = 50^\circ$, opposite side is $AC$)
• Adjacent to $50^\circ$: $BC = a$

Step3: Apply sine formula

Using $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$, for $\theta = 50^\circ$, we have:
$\sin(50^\circ)=\frac{3}{c}$ (because opposite side to $50^\circ$ is $3$ and hypotenuse is $c$).
Check other options:

• $\sin(50^\circ)=\frac{c}{3}$: Incorrect, as $\sin$ should be opposite over hypotenuse, not hypotenuse over opposite.
• $\cos(50^\circ)=\frac{c}{3}$: Incorrect, $\cos$ is adjacent over hypotenuse, and $c$ is hypotenuse, $3$ is opposite, not adjacent.
• $\cos(50^\circ)=\frac{3}{c}$: Incorrect, $3$ is opposite, not adjacent to $50^\circ$.

Answer:

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