Question

ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = ${110}^0$ and angle BAC = ${50}^0$. Select the statements that are true.

• A. $\angle DAC={60}^o$
• B. $\angle DCA={20}^o$
• C. $\angle DCA={10}^o$
• D. $\angle DAC={50}^o$

Answer 1

Answer: (A), (C)

The correct options are
A

$\angle DAC={60}^o$

C

$\angle DCA={10}^o$

ABCD is a cyclic quadrilateral in which AD || BC

$\angle ADC={110}^0,\angle BAC={50}^0$

$\angle B+\angle D={180}^0$

(Sum of opposite angles of a cyclic quadrilateral.)

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

$\therefore \angle Bor\angle ABC={180}^0-{110}^0={70}^0$

Now, in $\Delta$ ABC,

$\angle BAC+\angle ABC+\angle ACB={180}^0$

$\Rightarrow {50}^0+{70}^0+\angle ACB={180}^0$

$\Rightarrow {120}^0+\angle ACB={180}^0$

$\therefore \angle ACB={180}^0-{120}^0={60}^0$

$\because$ AD|| BC (Given)

$\therefore \angle DAC=\angle ACB$ (Alternate angles)

=${60}^0$

Now, in $\Delta$ ADC,

$\angle DAC+\angle ADC+\angle DCA={180}^0$

${60}^0+{110}^0+\angle DCA={180}^0$

${170}^0+\angle DCA={180}^0$

$\therefore \angle DCA={180}^0-{170}^0={10}^0$