Question

In circle O, PA and PB are tangents. The figure is not drawn to scale.

a. Prove that triangle APO is congruent to BPO


b. Find angle BOD for angle AOP = 64. Explain your reasoning.

Answer 1

a. ∡AOP = ∡BOP bounded by AO = BO and common segment OP

b. ∡BOD is 116°

Explanation:

Here, we have;

Line AO = BO → Radius of circle

∡OAP = ∡OBP = 90° →Tangent angle to a circle

Therefore, AP = BP = √((OP)² - (BO)²) = √((OP)² - (AO)²)

Hence since AO = BO and AP = BP as well as ∡OBP = ∡OAP, which are two sides and an included angle to completely define a triangle, we have that triangles APO is congruent to triangle BPO.

b. Here we have ∡AOP = 64°

Where, triangles APO and BPO are congruent, therefore ∡AOP = ∡BOP since the adjacent sides to the angles, AO and OP are equal to BO and OP

∴ ∡BOP = ∡AOP = 64°

Angle ∡BOD + ∡BOP = 180° = sum of angles on a straight line

Therefore, ∡BOD + 64° = 180° and ∡BOD = 180 - 64 = 116°

∡BOD = 116°.