Question

In the given figure, $ABCD$ is a cyclic quadrilateral in which $\angle CAD={25}^o,\angle ABC={50}^o$ and $\angle ACB={35}^o$.

Then: (i) $\angle CBD$ (ii) $\angle DAB$ (iii) $\angle ADB$ are respectively?

• A. (i) $\angle CBD={35}^o$
  (ii) $\angle DAB={80}^o$
  (iii) $\angle ADB={85}^o$
• B. (i) $\angle CBD={25}^o$
  (ii) $\angle DAB={35}^o$
  (iii) $\angle ADB={85}^o$
• C. (i) $\angle CBD={25}^o$
  (ii) $\angle DAB={70}^o$
  (iii) $\angle ADB={35}^o$
• D. (i) $\angle CBD={85}^o$
  (ii) $\angle DAB={70}^o$
  (iii) $\angle ADB={95}^o$

Answer 1

The correct option is C (i) $\angle CBD={25}^o$

(ii) $\angle DAB={70}^o$
(iii) $\angle ADB={35}^o$
Given $ABCD$ is a cyclic quadrilateral.

$AD$ & $BC$ have been joined.

$\angle CAD={25}^o,\angle ACB={35}^o,\angle ABC=50^o.$

$\angle CBD$ & $\angle CAD$ are angles subtended by the chord $CD$ to the circumference.

$\therefore \angle CBD=\angle CAD={25}^o$...[since the angles, subtended by a chord of a circle to the circumference of the same circle, a are equal].

Similarly, $\angle ACB$ & $\angle ADB$ are angles subtended by the chord $AB$ to the circumference.

$\therefore \angle ACB=\angle ADB={35}^o$ ...[since the angles, subtended by a chord of a circle to the circumference of the same circle, are equal].

Also, $\angle ADC$ & $\angle ABC$ are angles subtended by the chord $AB$ to the circumference.

$\therefore \angle ADC=\angle ABC={50}^o$...[since the angles, subtended by a chord of a circle to the circumference of the same circle, are equal].

$\therefore \angle BDC=\angle ADC+\angle ADB={50}^o+{35}^o={85}^o.$

Now $ABCD$ is a cyclic quadrilateral.

$\therefore$ The sum of its opposite angles $={180}^o.$

So $\angle BAC+\angle BDC={180}^o$

$⟹\angle BAC={180}^o-\angle BDC={180}^o-{85}^o={95}^o.$

$\therefore \angle DAB=\angle BAC-\angle CAD={95}^o-{25}^o={70}^o.$

So, (i) $\angle CBD={25}^o$, (ii) $\angle DAB={70}^o$, (iii) $\angle ADB={35}^o$.

Hence, Option $C$ is correct.