Question 57
The sides AB and AC of $$\triangle$$ABC are produced to points D and E, respectively. The bisectors of $$\angle$$CBD and $$\angle$$BCE meet at P If $$\angle$$A = $$72^\circ$$, then the measure of $$\angle$$PÂ is:
Solution
When the bisectors of $$\angle$$CBD and $$\angle$$BCE meet at P, then $$\angle$$P = $$90^{\circ\ }-\frac{\angle\ BAC}{2}$$
=Â $$90^{\circ\ }-\frac{72^{\circ\ }}{2}$$
= $$90^{\circ\ }-36^{\circ\ }$$
= $$54^{\circ\ }$$
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
