Question

In a cyclic trapezium ABCD, $\angle A={100}^{\circ }$. Then, $\angle D$ is equal to:

• A. ${80}^{\circ }$
• B. ${90}^{\circ }$
• C. ${100}^{\circ }$
• D. ${110}^{\circ }$

Answer 1

The correct option is A ${80}^{\circ }$

Consider given trapezium $ABCD$.

Given, $\angle A={100}^{\circ }$.

Now,

$\angle A=\angle B$ ....[Since, $DC$ is the diameter of the circle]

$⟹$ $\angle B={100}^{\circ }$.

Since, trapezium $ABCD$ is a cyclic quadrilateral,

Sum of opposite angle of trapezium $={180}^{\circ }$....[Opposite angles of cyclic quadrilateral are supplementary].

$⟹\angle B+\angle D={180}^{\circ }$

${100}^{\circ }+\angle D={180}^{\circ }$

$\angle D={80}^{\circ }$.

Hence, option $A$ is correct.