Can someone Answer this please?
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Answer:
The difference between the greatest and the smallest angle of the triangle PBQ is 98°.
Step-by-step explanation:
The question is:
In a triangle ABC, angle B measures 64 ° and angle C measures 72°. The inner bisector CD intersects the height BH and the bisector BM at P and Q respectively. Find the difference between the greatest and the smallest angle of the triangle PBQ.
Solution:
Consider the triangle ABC.
The measure of angle A is:
angle A + angle B + angle C = 180°
angle A = 180° - angle B - angle C
= 180° - 72° - 64°
= 44°
It is provided that CD and BM are bisectors.
That:
angle BCP = angle PCH = 36°
angle CBQ = angle QBD = 32°
angle BHC = 90°
Compute the measure of angle HBC as follows:
angle HBC = 180° - angle BHC + angle BCH
= 180° - 90° - 72°
= 18°
Compute the measure of angle BPC as follows:
angle BPC = 180° - angle PCB + angle CBP
= 180° - 18° - 36°
= 126°
Then the measure of angle BPQ will be:
angle BPQ = 180° - angle BPC
= 180° - 126°
angle BPQ = 54°
Compute the measure of angle PBQ as follows:
angle PBQ = angle B - angle QBD - angle HBC
= 64° - 32° - 18°
angle PBQ = 14°
Compute the measure of angle BQP as follows:
angle BQP = 180° - angle PBQ - angle BPQ
= 180° - 14° - 54°
angle BQP = 112°
So, the greatest and the smallest angle of the triangle PBQ are:
angle BQP = 112°
angle PBQ = 14°
Compute the difference:
<em>d</em> = angle BQP - angle PBQ
= 112° - 14°
= 98°
Thus, the difference between the greatest and the smallest angle of the triangle PBQ is 98°.
