Krishna
0

Step 1: Draw the diagram according to the given instructions.

NOTE: We are given a circle with centre O, an external point A and two

tangents AP and AQ to the circle, where P, Q are the points of contact.

Step 2: Prove that ∆APQ is an isosceles triangle

NOTE: AP = AQ, So ∆APQ is an isosceles triangle

Step 3: Calculate the two equal angles of Isosceles triangle  (∠APQ + ∠AQP)

NOTE: Sum of the interior angles in triangle = 180

EXAMPLE:  ∠APQ + ∠AQP + ∠PAQ = 180°

\angle APQ + \angle AQP = \frac{1}{2} (180 - \theta)  (since \angle PAQ = \theta)

\angle APQ + \angle AQP = 90 - \frac{1}{2}(\theta)

Step 4: Calculate the \angle OPQ of the isosceles triangles(∆OPQ) formed by the radiuses.

\angle OPQ = \angle OPA - \angle APQ

\angle OPQ = 90 - (90 - \frac{1}{2}\theta)

\angle OPQ = \frac{1}{2}\theta)

\angle OPQ = \frac{1}{2}\angle PAQ)

\angle PAQ = 2 * \angle OPQ

Similarly \angle PAQ = 2 * \angle OQP