Question

In the given construction of a $\Delta ABC$, $\angle CAB$ will be

• A. ${55}^{\circ }$
• B. ${45}^{\circ }$
• C. ${65}^{\circ }$

Answer 1

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


In $\Delta AXB$
PQ is the perpendicular bisector of XA, the two triangles formed are congruent by SAS rule.
$\therefore$ AB = XB
$\Rightarrow$ $\angle BAX=\angle AXB$

$\angle ABC=\angle BAX+\angle AXB$
= $2\angle AXB=\angle LXY$
[Since $\angle AXB$ is angle bisector of $\angle LXY$]

$Similarly,\angle ACB=\angle MYXTherefore,\angle ABC={75}^{\circ }and\angle BCA={60}^{\circ }In\Delta ABC,wehave\angle ABC+\angle BCA+\angle CAB={180}^{\circ }{75}^{\circ }+{60}^{\circ }+\angle CAB={180}^{\circ }\angle CAB={180}^{\circ }-{135}^{\circ }\angle CAB={45}^{\circ }$