Question

Given:

m∠QPS=107∘

m∠RPS=4x+27∘

m∠QPR=9x-115∘

Find m∠QPR


**This was on Khan Academy, so please hurry lol**

Answer 1

m∠QPR = 20°

Explanation:

If we make a sketch of the triangle, it will be observed, that line R bisects angle P.

Considering Triangle, ΔQPS;

∠QPS=107∘

Considering triangle, QPR;

∠QPR=9x-115∘

Considering triangle, RPS;

∠RPS=4x+27∘

Thus, ∠RPS + ∠QPR = ∠QPS

4x+27° + 9x-115° = 107°

13x - 88° = 107∘

13 x = 107∘ + 88∘

13x = 195°

x = 195°/13

x = 15°

m∠QPR = 9x - 115°

= (9 x 15) - 115°

= 135° - 115°

= 20°

Therefore, m∠QPR = 20°