Question

In the given construction,

Then $\angle CAB$ will be

• A. ${55}^{\circ }$
• B. ${45}^{\circ }$
• C. ${75}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

From construction
AB = XB (Since B is a point on the perpendicular bisector of AX)
$\Rightarrow \angle BAX=\angle AXB$ (angles opposite to equal sides are equal)

$\angle ABC=\angle BAX+\angle AXB$
= $2\angle AXB=\angle LXY$ ( $\because$ XA is the bisector of the $\angle LXY$ )

Similarly, $\angle ACB=\angle MYX$

Therefore, $\angle ABC={75}^{\circ }$ and $\angle BCA={60}^{\circ }$
In $△$ ABC, we have,
$\angle ABC+\angle BCA+\angle CAB={180}^{\circ }$
${75}^{\circ }+{60}^{\circ }+\angle CAB={180}^{\circ }$
Then $\angle CAB={180}^{\circ }-{135}^{\circ }$
Therefore $\angle CAB={45}^{\circ }$