In a figure, DE II QR, AP and BP are bisectors of angle EAB and angle RBA respectively. Find angle APB.
URGENT!!!
Attachments:
Sazz:
Still not, can't read it :(
Answers
Answered by
379
see diagram enclosed
Let angle EAP = angle BAP = x (AP bisector)
Let angle RBP = angle PBA = y BP is bisector
angle EAB + angle ABR = 180 degrees as AE and BR are parallel
2 * x + 2 * y = 180 deg x + y = 90
angle APB = 180 - x - y in trianle APB
angle APB = 180 - 90 = 90 deg
Let angle EAP = angle BAP = x (AP bisector)
Let angle RBP = angle PBA = y BP is bisector
angle EAB + angle ABR = 180 degrees as AE and BR are parallel
2 * x + 2 * y = 180 deg x + y = 90
angle APB = 180 - x - y in trianle APB
angle APB = 180 - 90 = 90 deg
Attachments:
Answered by
152
Step-by-step explanation:
Hope this works for everyone searching for this answer...
Attachments:
Similar questions