Question

The given figure shows a triangle ABC with angle BAC
= 56° and angle ABC = 64º. Bisectors of angles A, B
and C meet the circumcircle of the A ABC at points
P, Q and R respectively. Find the measure of
angle QPR.​

Answer 1

Answer:

Given△ABCisinscribedinC(0,r).

Thebisectorsof∠BAC,∠ABCand∠ACBmeetsthecircumcircle

of△ABC,inP,Q,Rrespectively.

InthefigureJoinRQ,

∠ABQ=∠APQ−(i)

{Anglesinthesamesegmentofacircleareequal}.

∠ABQ=∠QBC{BQisthebisectorof∠ABC}.

∴∠QBC=∠APQ−(ii)

Adding(i)&(ii)

∠ABQ+∠QBC=∠APQ+∠APQ

∴∠ABC=2∠APQ−(iii)

Similarily,∠ACB=2∠APR−(iv)

Adding(iii)&(iv)

∠ABC+∠ACB=2(∠APQ+∠APR)

∴∠ABC+∠ACB=2∠QPR−(v)

In△ABC,

∠ABC+∠BAC+∠ACB=180

∘

{Anglesumproperty}

∠ABC+∠ACB=180

∘

−∠BAC−(vi)

From(v)&(vi),weget

180

∘

−∠BAC=2∠QPR

∴∠QPR=

2

1

(180−∠BAC)

∠QPR=

2

1

×180

∘

−

2

1

∠BAC

⟹∠QPR=90−

2

1

∠BAC

Answer 2

Hy Dude ❣

Plz Refer the pic

Hope it helps you

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