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 $\angle$APY.s

Answer 1

In $\Delta$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 = ${30}^0$ (angle formed using the same chord on same segment)

and $\angle$ACX = $\angle$BXY +$\angle$ABX

= ${30}^0+{70}^0={100}^0$ (exterior angle of a triangle is the sum of the interior angle of the remaining two sides)

$\angle$ACY = $({180}^0-{100}^0)={80}^0$(Linear Pair)

$\angle$XYP =${90}^0$ [Diameter perpendicular to tangent]

In $\Delta$PCY

$\angle$APY = ${180}^0$ -$\angle$PCY - $\angle$CYP (Sum of the angles of triangles is ${180}^0$)

= ${180}^0-{80}^0-{90}^0={10}^0$