In Fig. 1, PQ and PR are tangents to the circle with centre O such that ∠QPR = 50°,
(a) 25°
(b) 30°
(c) 40°
(d) 50°
We are given the below figure in which
PQ and PR are tangents to the circle with centre O and
We have to find
PQ is the tangent to circle
Therefore [Since Radius of a circle is perpendicular to tangent]
Similarly
We know that sum of angles of a quadrilateral
Therefore in Quadrilateral PQOR
$\angle Q P R+\angle P R O+\angle R O Q+\angle O Q P=360^{\circ}$
$50^{\circ}+90^{\circ}+\angle R O Q+90^{\circ}=360^{\circ}$
$\angle R O Q=360^{\circ}-230^{\circ}$
$\angle R O Q=130^{\circ}$
In $\Delta O Q R$
$O Q=O R$ [Radius of same circle]
Therefore $\triangle O Q R$ is isosceles triangle
Hence $\angle \mathrm{ORQ}=\angle \mathrm{OQR}$ [Angles opposite to equal sides of isosceles triangle]
In $\triangle O Q R$
$\angle R O Q+\angle O Q R+\angle Q R O=180^{\circ}\left[\right.$ Sum of angles of a triangle $\left.=180^{\circ}\right]$
$2 \angle O Q R+130^{\circ}=180^{\circ}$
$\angle O Q R=\frac{50^{\circ}}{2}$
$=25^{\circ}$
Hence option (a) is correct.
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