Question

which equation can be solved to find one of the missing side lengths in the triangle?
options:
\\(\cos(60^\circ) = \frac{12}{a}\\)
\\(\cos(60^\circ) = \frac{12}{b}\\)
\\(\cos(60^\circ) = \frac{b}{a}\\)

Explanation:

Step1: Recall cosine definition

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

Step2: Identify sides for $\angle B = 60^\circ$

For $\angle B = 60^\circ$, adjacent side is $a$, hypotenuse is 12? Wait, no—wait, hypotenuse is the side opposite right angle, so hypotenuse is $AB = 12$? Wait, no, in triangle $ABC$, right-angled at $C$, so hypotenuse is $AB = 12$. Then for $\angle B = 60^\circ$, adjacent side is $BC = a$, hypotenuse is $AB = 12$? Wait, no, cosine of $\angle B$: adjacent over hypotenuse. Wait, no, let's re-express. Wait, maybe I mixed up. Wait, in right triangle at $C$, so sides: $BC = a$ (adjacent to $\angle B$), $AC = b$ (opposite to $\angle B$), hypotenuse $AB = 12$. So $\cos(60^\circ)=\frac{\text{adjacent to } \angle B}{\text{hypotenuse}}=\frac{a}{12}$? Wait, no, the options have $\cos(60^\circ)=\frac{12}{a}$, $\cos(60^\circ)=\frac{12}{b}$, etc. Wait, maybe I got the hypotenuse wrong. Wait, no—wait, maybe the hypotenuse is not 12? Wait, no, the side labeled 12 is $AB$, which is opposite the right angle at $C$, so $AB$ is hypotenuse, length 12. Then for $\angle B = 60^\circ$, adjacent side is $BC = a$, hypotenuse is $AB = 12$. So $\cos(60^\circ)=\frac{a}{12}$? But the options have $\cos(60^\circ)=\frac{12}{a}$, which would be if adjacent is 12 and hypotenuse is $a$, but that would mean $a$ is hypotenuse. Wait, maybe I mixed up the labels. Wait, maybe the side labeled 12 is not the hypotenuse? Wait, no, right angle at $C$, so hypotenuse is $AB$, so $AB = 12$. Then $\angle B = 60^\circ$, so adjacent to $\angle B$ is $BC = a$, hypotenuse $AB = 12$. So $\cos(60^\circ)=\frac{a}{12}$ → $a = 12\cos(60^\circ)$. But the first option is $\cos(60^\circ)=\frac{12}{a}$, which would imply $a = \frac{12}{\cos(60^\circ)}$, which would be if adjacent is 12 and hypotenuse is $a$. Wait, maybe the diagram has $AB = 12$, but maybe I got the adjacent and hypotenuse reversed. Wait, no—wait, maybe the hypotenuse is $a$? No, hypotenuse is opposite right angle, so must be $AB$. Wait, maybe the problem has a typo, or I misread. Wait, let's check the options again. The first option: $\cos(60^\circ)=\frac{12}{a}$. Let's solve for $a$: $a = \frac{12}{\cos(60^\circ)}$. If $\cos(60^\circ)=0.5$, then $a = 24$. But according to cosine definition, if $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$, then if $\cos(60^\circ)=\frac{12}{a}$, that would mean adjacent is 12, hypotenuse is $a$. So that would mean $a$ is hypotenuse, and 12 is adjacent to $\angle B$. So maybe the hypotenuse is $a$, and the side labeled 12 is adjacent to $\angle B$. So in that case, $\cos(60^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{12}{a}$. That must be the case. So the first option is correct.

Answer:

$\cos(60^\circ) = \frac{12}{a}$ (the first option)