Question

in △opq, q = 590 cm, m∠p = 152° and m∠q = 26°. find the length of p, to the nearest centimeter.

Explanation:

Step1: Find angle O

The sum of angles in a triangle is 180°. So, $m\angle O=180^{\circ}-m\angle P - m\angle Q$.
$m\angle O = 180^{\circ}-152^{\circ}-26^{\circ}=2^{\circ}$

Step2: Use the Law of Sines

The Law of Sines states that $\frac{p}{\sin P}=\frac{q}{\sin Q}$. We want to find $p$, so $p=\frac{q\sin P}{\sin Q}$.
Substitute $q = 590$ cm, $m\angle P = 152^{\circ}$, and $m\angle Q=26^{\circ}$ into the formula.
$p=\frac{590\times\sin(152^{\circ})}{\sin(26^{\circ})}$
We know that $\sin(152^{\circ})\approx0.4695$ and $\sin(26^{\circ})\approx0.4384$.
$p=\frac{590\times0.4695}{0.4384}=\frac{277.005}{0.4384}\approx632$ cm

Answer:

632 cm