Question

In the given figure, XY is a diameter of the circle, PQ is a tangent to the circle at Y. Given that $\angle$AXB = ${50}^0$ and $\angle$ABX = ${70}^0$, calculate $\angle$BAY

• A. 30°
• B. 40°
• C. 50°
• D. 60°

Answer 1

The correct option is A

30°

$\Delta$AXB,

$\angle$XAB + $\angle$AXB + $\angle$ABX = ${180}^0$

$\angle$XAB = ${60}^0$

$\angle$XAY = ${90}^0$ (Angles of semicircle)

$\angle$BAY = $\angle$XAY - $\angle$XAB

$={90}^0–{60}^0={30}^0$

$\angle$BXY = $\angle$BAY = ${30}^0$