Question

In the figure, if AOB is a diameter and $\angle ADC={120}^O$, then $\angle CAB={30}^O.$

State True or False.

• A. Cannot be determined.
• B. False.
• C. True.
• D. Sometimes false.

Answer 1

Answer: (C)

True.

$m\angle ADC=\frac{1}{2}m\left(arcABC\right)$

Therefore, $m\left(arcABC\right)={240}^o$

Now, $m\left(arcABC\right)+m\left(arcADC\right)={360}^o$

Therefore, $m\left(arcADC\right)=360-240={120}^o$

Now, Construct $△ABC$

Hence, $m\left(\angle ABC\right)$$=\frac{1}{2}m\left(arcADC\right)$

Therefore, $m\left(\angle ABC\right)$$={60}^o$

Now, from the new construction,

$m\angle ABC+m\angle CAB+m\angle ACB={180}^o$

Therefore, $m\angle CAB={30}^o$