Solve for T Leave your answer in simplest radical form.
1 answer:
You might be interested in
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°.
If <em>X</em> is uniformly distributed on the interval (0, 12), then its PDF is
or simply
and the zero-elsewhere case is assumed.
Whether you include 0 and 12 in the domain is irrelevant, since the probability that <em>X</em> = 0 or <em>X</em> = 12 are both zero.
Answer:
I love algebra anyways
The ans is in the picture with the steps how i got it
(hope this helps can i plz have brainlist :D hehe)
Step-by-step explanation:
