Question

in $\triangle pqr$, $sin r=\frac{3}{5}$. what is $cos p$?

Explanation:

Step1: Calculate hypotenuse PR

$$PR = \sqrt{PQ^2 + QR^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20$$

Step2: Find cos P

$$\cos P = \frac{\text{adjacent to } P}{\text{hypotenuse}} = \frac{PQ}{PR} = \frac{12}{20} = \frac{3}{5}$$

Answer:

$\frac{3}{5}$