Question

In the given figure, C and D are points on the semi-circle described on AB as diameter. Given angle BAD = ${70}^0$ and angle DBC= ${30}^0$. Calculate $\angle BDC$

• A. $\angle BDC={40}^o$
• B. $\angle BDC={70}^o$
• C. $\angle BDC={10}^o$
• D. $\angle BDC={50}^o$

Answer 1

The correct option is A

$\angle BDC={40}^o$

$\therefore$ ABCD is a cyclic quadrilateral.

$\angle BAD+\angle BCD={180}^0$ (Sum of opposite angles)

$\Rightarrow {70}^0+\angle BCD={180}^0$

$\Rightarrow \angle BCD={180}^0-{70}^0={110}^0$

Now in $\Delta BCD$.

$\angle BCD+\angle DBC+\angle BDC={180}^0$

$\Rightarrow {30}^0+{110}^0+\angle BDC={180}^0$

$\Rightarrow {140}^0+\angle BDC={180}^0$

$\therefore \angle BDC={180}^0-{140}^0={40}^0$