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 $$\angle$$P is:
Solution
When the bisectors of $$\angle$$CBD and $$\angle$$BCE meet at P, then $$\angle$$P = $$90^{\circ\ }-\frac{\angle\ BAC}{2}$$
= $$90^{\circ\ }-\frac{72^{\circ\ }}{2}$$
= $$90^{\circ\ }-36^{\circ\ }$$
= $$54^{\circ\ }$$
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