Question

given triangle abc, which equation could be used to find the measure of ∠b? cos m∠b = \frac{\sqrt{5}}{5} sin m∠b = \frac{\sqrt{5}}{5} cos m∠b = \frac{\sqrt{5}}{2} sin m∠b = \frac{2\sqrt{5}}{5}

Explanation:

Step1: Recall trigonometric - ratio definitions

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$ for an acute angle $\theta$. For $\angle B$ in right - triangle $ABC$, the adjacent side to $\angle B$ is $AB = 4$, the opposite side to $\angle B$ is $AC = 2$, and the hypotenuse $BC=\sqrt{4^{2}+2^{2}}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}$.

Step2: Calculate $\sin B$ and $\cos B$

$\sin B=\frac{AC}{BC}=\frac{2}{2\sqrt{5}}=\frac{\sqrt{5}}{5}$, $\cos B=\frac{AB}{BC}=\frac{4}{2\sqrt{5}}=\frac{2\sqrt{5}}{5}$.

Answer:

$\sin m\angle B=\frac{\sqrt{5}}{5}$