Answer:
Step-by-step explanation:
given D : (7,-3), and D' : (2,5)
the coordinates of D can be represented as (x1,y1), and the coordinates of D' can be represented as (x,y).
you can simply take the difference in the x values and difference in the y values from the preimage to image.
like this:
f'(x,y) → f(x+(x-x1),y+(y-y1)) :
D'(x,y) → D(x+(2-7),y+(5--3))
D'(x,y) → D(x<u>-5</u>,y<u>+8</u>) :
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°.
