Q22 of 52 Page 10

In Fig. 2, O is the center of the circle and LN is a diameter. If PQ is a tangent to the circle at K and KLN = 30°, find PKL . (CBSE 2017 C)



Given: OLK = 30°

KO = OL (Represents the radius of the same circle)


So ΔOKL is isosceles


OKL = OLK = 30° (Base angles of an isosceles triangle)


OKP = 90° (Tangents are always perpendicular with the line joining the center)


PKL = OKP-OKL


⇒ PKL = 90°-30°


Answer: PKL = 60°

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