Question

In the given figure, $\angle ABD={90}^o,\angle BDA={30}^o$ and $\angle BCA={20}^o$. What is $\angle CAD$?

• A. ${10}^o$
• B. ${20}^o$
• C. ${30}^o$
• D. ${15}^o$

Answer 1

The correct option is A ${10}^o$

R.E.F image

In $\Delta ABD$

$x+30+90={180}^{\circ }$

$\Rightarrow x={60}^{\circ }$

In $\Delta ABC-(x+y)+20+{90}^{\circ }={180}^{\circ }$

$\Rightarrow y={20}^{\circ }-x\Rightarrow y={10}^{\circ }$