in fig seg pq is a tangent and op is a radius angle oqp=40°then find the measure of angle opq and poq
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Answer:
Given- O is the centre of a circle to which PQ is a tangent at P. ΔOPQ is isosceles whose vertex is P.
To find out- ∠OQP=?
Solution- OP is a radius through P, the point of contact of the tangent PQ with the given circle ∠OPQ=90
o
since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent. Now ΔOPQ is isosceles whose vertex is P.
∴OP=PQ⟹∠OQP=∠QOP⟹∠OQP+∠QOP=2∠OQP.
∴ By angle sum property of triangles,
∴∠OPQ+2∠OQP=180
o
⟹90
o
+2∠OQP=180
o
⟹∠OQP=45
o
.
Ans- Option- B.
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