In the figure. O is the centre of circle. P q is chord and pt is a tangent tonthe circle at p . Find tpq
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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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Answer:
In the given figure, O is the center of a circle, PQ is a chord and Pt is the tangent at P. If ∠POQ =70o , then ∠TPQ is equal to (a) 35o (b) 45o (c) 55o (d) 70oRead more on Sarthaks.com - https://www.sarthaks.com/158680/in-the-given-figure-the-center-circle-chord-and-is-the-tangent-at-if-poq-70-then-tpq-is equal to
(a) 35o (b) 45o (c) 55o (d) 70o
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