in given figure boa is a diameter of a circle and the tangent at a poitn p meet ba extended at t if angle pbo =30 then angle pta is equal to
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Given, PT is a tangent
∠PBA=30
0
Now,
∠BPA=90
0
(angle in semicircle)
∠OPT=90
0
(angle between the radius and tangent)
Also in ΔBPA
∠PAB=180
0
−(∠30
0
+∠BPA)
=180
0
−30
0
−90
0
=60
0
⟶(1)
Now look into the ΔOPA,
OP=OA (radius of circle)
⇒∠OAP=∠OPA=60
0
(∵∠OAP=∠BAP=60
0
)
∴ ∠APT=∠OPT−∠OPA=90
0
−60
0
=30
0
∴ ∠PAT=180
0
−(60
0
)=120
0
(supplementary angle)
In ΔPAT,
∠PAT+∠PTA+∠APT=180
0
⇒120
0
+∠PTA+30
0
=180
0
⇒∠PTA=30
0
.
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