Question

in △pqr, sin r = \frac{3}{5}. what is cos p? image of right triangle pqr with right angle at q, pq = 12, qr = 16, and multiple - choice options below

Explanation:

Step1: Identify triangle type

$\triangle PQR$ is right - angled at $Q$, so $\angle Q = 90^{\circ}$, and $\angle P+\angle R=90^{\circ}$. So $\cos P=\sin R$ (since for complementary angles $A$ and $B$ ($A + B=90^{\circ}$), $\cos A=\sin B$).

Step2: Use given $\sin R$

We are given that $\sin R=\frac{3}{5}$? Wait, wait, let's check the sides. Wait, $PQ = 12$, $QR = 16$. Let's calculate the hypotenuse $PR$ using Pythagoras theorem. $PR=\sqrt{PQ^{2}+QR^{2}}=\sqrt{12^{2}+16^{2}}=\sqrt{144 + 256}=\sqrt{400} = 20$. Now, $\sin R=\frac{PQ}{PR}=\frac{12}{20}=\frac{3}{5}$, which matches the given. Now, $\angle P$ and $\angle R$ are complementary. So $\cos P=\sin R$. Since $\sin R=\frac{3}{5}$? Wait, no, wait. Wait, $\cos P=\frac{PQ}{PR}$? Wait, no. Let's find $\cos P$. In right - triangle $PQR$, $\cos P=\frac{PQ}{PR}$. $PQ = 12$, $PR = 20$, so $\cos P=\frac{12}{20}=\frac{3}{5}$? Wait, no, wait. Wait, $\angle P$: the adjacent side to $\angle P$ is $PQ$, and the hypotenuse is $PR$. Wait, $PQ = 12$, $PR = 20$. So $\cos P=\frac{PQ}{PR}=\frac{12}{20}=\frac{3}{5}$. Alternatively, since $\angle P+\angle R = 90^{\circ}$, $\cos P=\sin R$. $\sin R=\frac{PQ}{PR}=\frac{12}{20}=\frac{3}{5}$, so $\cos P=\frac{3}{5}$. Wait, but let's check again. Wait, $PQ = 12$, $QR = 16$, $PR = 20$. For $\angle P$: adjacent side is $PQ = 12$, hypotenuse is $PR = 20$. So $\cos P=\frac{adjacent}{hypotenuse}=\frac{PQ}{PR}=\frac{12}{20}=\frac{3}{5}$.

Answer:

$\frac{3}{5}$