In Fig. 2, XAY is a tangent to the circle centered at O. If $\angle ABO = 40°$, then find $m\angle BAY$ and $…

Question

In Fig. 2, XAY is a tangent to the circle centered at O. If $\angle ABO = 40°$, then find $m\angle BAY$ and $m\angle AOB$.

Accepted Answer

## Model Answer Since OA is radius and XAY is tangent, **∠OAY = 90°** (radius ⊥ tangent). In △OAB, OA = OB (radii), so △OAB is isosceles. ∴ ∠OAB = ∠OBA = 40° ∠BAY = ∠OAY − ∠OAB = 90° − 40° = **50°** ∠AOB = 180° − 40° − 40° = **100°** Source: Chapter 10, Section 10.2 (Theorem 10.1) --- ## Explanation Two key properties are tested here: 1. **Radius ⊥ tangent** at point of contact → ∠OAY = 90°. 2. **OA = OB** (radii) → isosceles triangle → base angles equal → ∠OAB = ∠OBA = 40°. Then ∠BAY = 90° − 40° = 50° and ∠AOB = 180° − 80° = 100° (angle sum of triangle). Examiners expect both values clearly stated with brief reasoning.

Mathematics — Standard (041) — CBSE Class 10 board question