Question

hii guys tomorrow is my exam please solve this question​

Answer 1

so here is your answer.

it is given that PQ is a diameter ,so angle PXQ =90°

(angle on diameter)

and it is given that XY is equal to radius so,

OX=XY=OY

so ΔXOY is equilateral triangle

so angle OXY=Angle OYX =angle XOY =60°

PE is a straight line so ,angle PXQ+angleQXE=180°

but, angle PXQ=90°(proved above)

therefore, angle QXE =90°

XY is a arc ,

angle XOY=2angle XQY

angle XOY =60°(proved above)

so ,angle XQY=30°(angle subtended by arc at centre is twice angle subtended by it at remaining part of circle)

so in ΔXQE

angle XQE+angle QXE+angle XEQ=180°

30°+90°+angle XEQ=180°

120°+angle XEQ=180°

angleXEQ=180°-120°

angle XEQ=60°

or, angle PEQ=60°

Hope it will help you.