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