Question

reviewing the sine ratio. which sine ratios are correct for △pqr? check all that apply. sin (p) = \frac{r}{q}, sin (p) = \frac{p}{q}, sin (q) = \frac{r}{p}, sin (r) = \frac{q}{r}, sin (r) = \frac{r}{q}

Explanation:

Step1: Recall sine - ratio definition

In a right - triangle, the sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse. For right - triangle $\triangle PQR$ with right - angle at $Q$, the hypotenuse is $q$.

Step2: Calculate $\sin(P)$

The side opposite angle $P$ is $r$. So, $\sin(P)=\frac{r}{q}$.

Step3: Calculate $\sin(Q)$

Since $\angle Q = 90^{\circ}$, $\sin(Q)=\sin(90^{\circ}) = 1$. But using the ratio formula, the side opposite $\angle Q$ is the hypotenuse $q$, so $\sin(Q)=\frac{q}{q}=1$. The given $\sin(Q)=\frac{r}{p}$ is incorrect.

Step4: Calculate $\sin(R)$

The side opposite angle $R$ is $p$. So, $\sin(R)=\frac{p}{q}$. The given $\sin(R)=\frac{q}{r}$ and $\sin(R)=\frac{r}{q}$ are incorrect.

Answer:

$\sin(P)=\frac{r}{q}$