Question

In the figure, AB is a diameter, $\angle$BAC$={38}^o$. Find $\angle$ADC.

Answer 1

Given that:

$\angle BAC={38}^{\circ }$ and AB is a diameter.

To find:

$\angle ADC=?$

Solution:

Since, AB is a diameter.

$\angle ACB={90}^{\circ }$

In $△ACB$

$\angle BAC+\angle ACB+\angle ABC={180}^{\circ }$

or, ${38}^{\circ }+{90}^{\circ }+\angle ABC={180}^{\circ }$

or, $\angle ABC={52}^{\circ }$

Now,

Since $ABCD$ is a cyclic quadrilateral.

$\angle ADC+\angle ABC={180}^{\circ }$ (sum of opposite angles of a cyclic quadrilateral is ${180}^{\circ }.$)

or, $\angle ADC+{52}^{\circ }={180}^{\circ }$

or, $\angle ADC={128}^{\circ }$