Question

in the given figure,PQ is a tangent draw from a point P to a circle with center O and QOR is a circle diameter of the circle such that​

Answer 1

Answer:

Correct option is

C

30

o

Given- PQ is a tangent to a circle with centre O at Q. QOR is a diameter of the given circle so that ∠POR=120

o

. To find out- ∠OPQ=?

Solution- QOR is a diameter.

∴OQ is a radius through the point of contact Q of the tangent PQ. ∴∠OQP=90

o

since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent.∴∠OPQ+∠OQP=120

o

(external angles of a triangle=sum of the internal opposite angles )

∴∠OPQ=120

o

−90

o

=30

o

.

Ans- Option C.