Question

In the given figures, O is the centre of the circle and AB is a tangent to it at point B. $\angle$BDC = ${65}^0$ Find $\angle$BAO.

• A. 40°
• B. 60°
• C. 30°
• D. 50°

Answer 1

The correct option is A

40°

Let, $\angle$BAO = $x^0$

$\angle$ABO = ${90}^0$ [$\because$ OB$\perp$ AB]

In $\Delta$BDC,

$\angle$B + $\angle$D + $\angle$C = ${180}^0$(sum of the angles of triangle)

$\angle$C = ${180}^0-{90}^0-{65}^0={25}^0$

Now, $\angle$BOE = 2$\angle$C = ${50}^0$

So, In $\Delta$ABO,

$\angle$A + $\angle$B + $\angle$O = ${180}^0$

$\angle$A = ${180}^0-{90}^0–{50}^0={40}^0$

So, $\angle$BAO = ${40}^0$