Question

In the figure, AB is a diameter of a circle with centre O and CD. If $\angle$BAC = ${20}^{\circ }$, find the values of (i)$\angle$BOC (ii)$\angle$DOC (iii)$\angle$DAC (iv)$\angle$ADC

Answer 1

1. Arc BC subtends $\angle$BOC at the centre and $\angle$BAC at the circumference.

$\angle$BOC = 2$\angle$BAC = (2 x ${20}^{\circ }$) = ${40}^{\circ }$. [Since angle at the centre is double the

angle at circumference]

1. $\angle$OCD = $\angle$BOC = ${40}^{\circ }$ [Alt. interior angles as CD || BA]

Now, OC = OD [Radii of the same circle]

$\angle$ODC = $\angle$OCD =${40}^{\circ }$

In $△$OCD, we have

$\angle$DOC + $\angle$OCD + $\angle$ODC =${180}^{\circ }$ (Sum of the angles of a triangle is ${180}^{\circ }$)

$\angle$DOC + ${40}^{\circ }$ + ${40}^{\circ }$ = ${180}^{\circ }$

$\angle$DOC = (${180}^{\circ }$ - ${80}^{\circ }$) = ${100}^{\circ }$

1. $\angle$DAC = $\frac{\angle \text{DOC}}{2}$= ${50}^{\circ }$ [Angle at the centre is double the angle at circumference]
2. $\angle$ACD = $\angle$CAB = ${20}^{\circ }$ [alternate interior angles, as CD || BA]

In $△$ACD, we have

$\angle$ADC + $\angle$ACD + $\angle$DAC =${180}^{\circ }$ [Sum of the angles of a triangle is ${180}^{\circ }$]

Hence, $\angle$BOC = ${40}^{\circ }$, $\angle$DOC = ${100}^{\circ }$, $\angle$DAC = ${50}^{\circ }$ and $\angle$ACD = ${110}^{\circ }$