Question

In the adjoining figure, angle subtended by the arc $\text{AB}$ at the center of smaller and bigger cirles are ${60}^{\circ }$ and ${40}^{\circ }$ respectively, then $\angle \text{CAD}$ is

• A. ${80}^{\circ }$
• B. ${130}^{\circ }$
• C. ${120}^{\circ }$
• D. ${100}^{\circ }$

Answer 1

The correct option is B ${130}^{\circ }$


Arc $\text{AB}$ makes ${60}^{\circ }$ at the center of the smaller circle. So, $\angle \text{ADB}$ will be half of ${60}^{\circ }$
i.e. $\angle \text{ADB}=\frac{{60}^{\circ }}{2}$
$={30}^{\circ }$
$\Rightarrow \angle \text{ADB}=\angle \text{ADC}={30}^{\circ }$

Arc $\text{AB}$ makes ${40}^{\circ }$ at the center of the bigger circle. So, $\angle \text{ACB}$ will be half of ${40}^{\circ }$
i.e. $\angle \text{ACB}=\frac{{40}^{\circ }}{2}$
$={20}^{\circ }$
$\Rightarrow \angle \text{ADB}=\angle \text{ACD}={20}^{\circ }$
Now,
$\angle \text{ACD}+\angle \text{CDA}+\angle \text{DAC}={180}^{\circ }$
${20}^{\circ }+{30}^{\circ }+\angle \text{DAC}={180}^{\circ }$
$\angle \text{DAC}={180}^{\circ }-{50}^{\circ }$
$\angle \text{DAC}={130}^{\circ }$
Thus, $\angle \text{DAC}$ or $\angle \text{CAD}$ is ${130}^{\circ }$.