Question

In the given figure, AD and CD are the chords of the circle with centre 'O'. If AD = CD, ∠AOB = 60° and ∠BOC = 30°, then find ∠ADC.

Answer 1

We know that equal chords of a circle subtend equal angles at the centre.

$\Rightarrow \angle AOD=\angle COD=x$

Since, the central angle of a circle is ${360}^{\circ }$

So, $\angle AOB+\angle BOC+\angle COD+\angle AOD={360}^{\circ }$

$\Rightarrow {60}^{\circ }+{30}^{\circ }+x+x={360}^{\circ }$

$\Rightarrow {90}^{\circ }+2x={360}^{\circ }$
$\Rightarrow 2x={270}^{\circ }$
$\Rightarrow x={135}^{\circ }$
∠COD = ∠AOD = ${135}^{\circ }$
Consider triangle COD.
CO = OD (radii of the circle)
∠OCD = ∠ODC (angles opposite to the equal sides)
Assume, ∠OCD = ∠ODC = y
Use angle sum property of the triangle.
∠OCD + ∠ODC + ∠COD = ${180}^{\circ }$
$\Rightarrow y+y+{135}^{\circ }={180}^{\circ }$
$\Rightarrow 2y={45}^{\circ }$
$\Rightarrow y={22.5}^{\circ }$
∠ODC = ${22.5}^{\circ }$
Similarly,
Consider triangle AOD.
AO = OD (radii of the circle)
∠OAD = ∠ODA (angles opposite to the equal sides)
Assume, ∠OAD = ∠ODA = z
Use angle sum property of the triangle.
∠OAD + ∠ODA + ∠AOD = ${180}^{\circ }$
$\Rightarrow z+z+{135}^{\circ }={180}^{\circ }$
$\Rightarrow 2z={45}^{\circ }$
$\Rightarrow z={22.5}^{\circ }$
∠ODA = ${22.5}^{\circ }$
∠ADC = ∠ODC + ∠ODA
∠ADC = ${22.5}^{\circ }$ + ${22.5}^{\circ }$
∠ADC = ${45}^{\circ }$