in the given figure, AB is a tangent to the circle in a point P. if angle=150DEEGREE then find angle OQP
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Answer:
OP≅OQ(Radii of the same circle)
∴ΔPOQ is an isosceles triangle
∴∠OPQ≅∠OQP ……………(1)
Let ∠OPQ⋅∠OQP=x ………………(2)
In ΔOPQ,
∠OPQ+∠OQP+∠POQ=180
o
x+x+70
o
=180
o
2x=180
o
−70
o
2x=110
o
x=110
o
/2
∴x=55
o
PT is the tangent to the circle at P.
∴OP⊥PT
(tangent is perpendicular to radius)
∴∠OPT=90
o
∠OPT=∠OPQ+∠TPQ
90
o
=55
o
+∠TPQ
⇒∠TPQ=90
o
−55
o
∠TPQ=35
o
.
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