Question

In the following figure, XAY is a tangent to the circle centered at O. If ∠ABO = 40°, then find m∠BAY  and m∠AOY.

Answer 1

Given: XAY is a tangent at A to the circle with centre O, and ∠ABO = 40°.

Step-wise calculation:

1. OA = OB (radii), so triangle AOB is isosceles.

Hence ∠OAB = ∠ABO = 40°.   ...(Base angles of isosceles triangle)

2. ∠AOB = 180° – (∠OAB + ∠ABO)

= 180° – (40° + 40°)

= 100°

3. The tangent at A is perpendicular to the radius OA, so ∠OAY = 90°.

4. ∠BAY = ∠OAY – ∠OAB   ...(Angle between tangent and chord = Complement of angle between chord and radius at A)

= 90° – 40°

= 50°   

5. From the figure Y lies on the line OB produced, so ∠AOY = ∠AOB = 100°. (OY is the extension of OB, so the central angle between OA and OY equals ∠AOB computed above).

m∠BAY = 50° and m∠AOY = 100°.