can anyone solve it??
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Answered by
2
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
So OPB is 90
BPQ = 90 - OPB
Angle on semicircle is 90
RQP =90
OPQ + ORQ = 90
ORQ = 90 - OPQ
HENCE BPQ = PRQ
Neriah:
mark as brainliest
Answered by
4
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
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
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