Question

In the figure, $\angle BAD={70}^{\circ }$, $\angle ABD={56}^{\circ }$ and $\angle ADC={72}^{\circ }$.

Calculate (i) $\angle BDC$ (ii) $\angle BCD$ (iii) $\angle CBD$.

• A. (i) $\angle BDC={18}^{\circ }$
  (ii) $\angle BCD={110}^{\circ }$
  (iii) $\angle CBD={52}^{\circ }$
• B. (i) $\angle BDC={18}^{\circ }$
  (ii) $\angle BCD={110}^{\circ }$
  (iii) $\angle CBD={54}^{\circ }$
• C. (i) $\angle BDC={18}^{\circ }$
  (ii) $\angle BCD={120}^{\circ }$
  (iii) $\angle CBD={52}^{\circ }$
• D. (i) $\angle BDC={10}^{\circ }$
  (ii) $\angle BCD={110}^{\circ }$
  (iii) $\angle CBD={52}^{\circ }$

Answer 1

The correct option is A (i) $\angle BDC={18}^{\circ }$

(ii) $\angle BCD={110}^{\circ }$
(iii) $\angle CBD={52}^{\circ }$
Given,
$\angle BAD={70}^{\circ }$, $\angle ABD={56}^{\circ }$ and $\angle ADC={72}^{\circ }$.

ABCD is a cyclic quadrilateral.

We know, the sum of opposite angles of a cyclic quadrilateral is ${180}^o$.
Therefore,
$\angle BAD+\angle BCD={180}^{\circ }$

$\angle BCD={180}^{\circ }-\angle BAD$
$\angle BCD={180}^{\circ }-{70}^{\circ }$ $={110}^{\circ }$ .....(i).

Also, $\angle ADC+\angle ABC={180}^o$
${72}^{\circ }+({56}^{\circ }+\angle CBD)={180}^{\circ }$
$\angle CBD={180}^{\circ }-{128}^{\circ }$ $={52}^{\circ }$ .....(ii).

In $△BCD$,
$\angle BCD+\angle DBC+\angle BDC={180}^{\circ }$ ...[Angle sum property]
${110}^{\circ }+{52}^{\circ }+\angle BDC={180}^{\circ }$
$\angle BDC={18}^{\circ }$ ...(i).

Therefore, option $A$ is correct.