Question

ABCD is a cyclic quadrilateral in which :

(i) BC || AD, $ADC={110}^{o}and\angle BAC={50}^o$. Find $\angle DAC$.

(ii) $\angle DBC={80}^{o}and\angle BAC={40}^o.$ Find $\angle BCD$.

(iii) $\angle BCD={100}^{o}andABD={70}^o$. Find $\angle ADB$.

Answer 1

(i) In the figure,

ABCD is a cyclic quadrilateral and AD || BC, $\angle ADC={110}^o$

$\angle BAC={50}^o$

$\because \angle B+\angle D={180}^o$ (Sum of opposite angles)

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

$\therefore \angle B={180}^o-{110}^o={70}^o$

Now in $\Delta ABC$,

$\angle CAB+\angle ABC+\angle BCA={180}^o$ (Sum of angles of a triangle)

$\Rightarrow {50}^o+{70}^o+\angle BCA={180}^o$

$\Rightarrow {120}^o+\angle BCA={180}^o$

$\Rightarrow BCA={180}^o-{120}^o={60}^o$

But $\angle DAC=\angle BCA$ (Alternate angles)

$\therefore \angle DAC={60}^o$

(ii) In cyclic quadrilateral ABCD, Diagonals AC and BD are joined.

$\angle DBC={80}^o,\angle BAC={40}^o$

Arc DC subtends $\angle DBCand\angle DAC$ in the same segment.

$\therefore \angle DBC=\angle DAC={80}^o$

$\therefore \angle DAB=\angle DAC+\angle CAB$$={80}^o+{40}^o={120}^o$

But $\angle DAB+\angle BCD={180}^o$ (Sum of opposite angles of a cyclic quadrilateral).

$\Rightarrow {120}^o+\angle BCD={180}^o$

$\Rightarrow \angle BCD={180}^o-{120}^o={60}^o$

(iii) In the figure, ABCD is a cyclic quadrilateral and BD is joined.

$\angle BCD={100}^o$ and $\angle ABD={70}^o$

$\angle A+\angle C={180}^o$ (Sum of opposite angles of cyclic quadrilateral)

$\Rightarrow \angle A+{100}^o={180}^o$

$\Rightarrow \angle A={180}^o-{100}^o$

$\therefore \angle A={80}^o$

Now in $\Delta ABD$,

$\angle A+\angle ABD+\angle ADB={180}^o$ (Sum of angles of a triangle)

$\Rightarrow {80}^o+{70}^o+\angle ADB={180}^o$

$\Rightarrow {150}^o+\angle ADB={180}^o$

$\Rightarrow \angle ADB={180}^o-{150}^o={30}^o$

$\therefore \angle ADB={30}^o$