Q7 of 47 Page 1

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.


(Fig. 2)

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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(Fig. 3)