You can get a circle by cutting horizontally and a hyperbola by cutting vertically. You can also get an ellipse and a parabola by cutting at an angle depending on whether the cut comes out the other side or not.
Find the area of the triangle below. Be sure to include the correct unit in your answer.
1 answer:
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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°.
