Question

En un triangulo ABC, el angulo B mide 64° y el angulo C mide 72°. La bisectriz interior CD corta a la altura BH y a la bisectriz BM en P y Q respectivamente. Hallar la diferencia entre el mayor y menor angulo del triangulo PBQ.

Answer 1

The difference between the greatest and the smallest angle of the triangle PBQ is 98°.

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:

d = angle BQP - angle PBQ

= 112° - 14°

= 98°

Thus, the difference between the greatest and the smallest angle of the triangle PBQ is 98°.