Question

can anyone solve it??

Answer 1

A tangent is perpendicular to Radius.
So OPB is 90
BPQ = 90 - OPB

Angle on semicircle is 90
RQP =90

OPQ + ORQ = 90
ORQ = 90 - OPQ

HENCE BPQ = PRQ

Answer 2

HELLO DEAR,




WE KNOW THAT THE AB IS A TANGENT,

THEN.

<BPR=90°

AND,

<BPQ + <RPQ =90 °

<RPQ =90°-<BPQ--------------(1)


<PQR=90° (TRIANGLE DRAWN ON

DIAMETER IS ALWAYS 90°)-----(2)

AND NOW,


IN ∆PQR
<PQR +<RPQ +<PRQ =180°

<PQR +90°-<BPQ +<PRQ =180° USING (1)


<PQR +<PRQ =180° -90° +<BPQ


<PQR +<PRQ = 90° + <BPQ--------

NOW ,

<PQR +90°= 90° +<BPQ .... USING (2)


THEN THE,

<PQR =<BPQ +90° -90°

<PQR = <BPQ


I HOPE ITS HELP YOU DEAR,
THANKS