Question

(d) problem 20: (first taught in lesson 30)
find m∠p and m∠pqc.
after you enter your answer press go.
m∠p =
m∠pqc =

Explanation:

Step1: Find $m\angle PQC$

Since an exterior - angle of a triangle is equal to the sum of the two non - adjacent interior angles and the exterior angle at $Q$ is $125^{\circ}$, and $\angle PQC$ and the $125^{\circ}$ angle are linear pairs. So $m\angle PQC=180 - 125=55^{\circ}$.

Step2: Find $m\angle P$

In right - triangle $PQC$ with $\angle C = 90^{\circ}$, using the angle - sum property of a triangle ($m\angle P+m\angle PQC + m\angle C=180^{\circ}$). Substitute $m\angle PQC = 55^{\circ}$ and $m\angle C = 90^{\circ}$ into the formula: $m\angle P=180-(90 + 55)=35^{\circ}$.

Answer:

$m\angle P = 35^{\circ}$, $m\angle PQC = 55^{\circ}$