Question

in an isocelese ∆PQR , the bisector of angle Q and angle R meet at O. if angle QOR = 140° then angle P =?​

Answer 1

Step-by-step explanation:

Given that

OQ and OR are angle bisector of Q and R

PQR is a triangle

also from figure OQR is a triangle

Let angle Q= 2x

angle R = 2y

Now in ∆ PQR

angle Q + angle R + angle P = 180°

or 2x + 2y +angle P = 180°

=> x+y = 90° - 1/2 angle P .......... (i)

Now in ∆OQR

angle RQO + angle QRO + angle O = 180

as angle RQO = x (from figure)

and angle QRO = y(from figure)

=> x + y + angle O = 180°

=> angle O = 180° - (x+y) ........(ii)

Now from (i)and (ii)

Angle O = 180° - ( 90° - 1/2 angle P )

=> angle O = 90° + 1/2 angle P

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Answer 2

See attachment.

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