Question

triangle abc is shown. what is sin(∠bac)? choose 1 answer: a $\frac{9sqrt{97}}{97}$ b $9sqrt{97}$ c $\frac{4sqrt{97}}{97}$ d $\frac{9}{4}$

Explanation:

Step1: Recall sine - definition

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. For $\angle BAC$, the side opposite to $\angle BAC$ is $BC = 9$, and the hypotenuse is $AB=\sqrt{97}$.

Step2: Calculate $\sin(\angle BAC)$

$\sin(\angle BAC)=\frac{BC}{AB}=\frac{9}{\sqrt{97}}$. Rationalize the denominator: $\frac{9}{\sqrt{97}}\times\frac{\sqrt{97}}{\sqrt{97}}=\frac{9\sqrt{97}}{97}$.

Answer:

A. $\frac{9\sqrt{97}}{97}$