Question

ABCDE...... is part of a polygon which has interior angles of ${160}^0$.CDLM is a square.
Find the value of x and y respectively.

• A. $70,105$
• B. $70,150$
• C. $105,70$
• D. $150,70$

Answer 1

The correct option is A $70,105$

$\left(i\right)\because CDLM$ is a square, $\angle DCM={90}^{\circ }.$

But $\angle BCD={160}^{\circ }$ (interiror angle of polygon)

$\Rightarrow \angle BCM=x^{\circ }=\angle BCD-\angle DCM={160}^{\circ }-{90}^{\circ }={70}^{\circ }\to \left(1\right)$

$\Rightarrow x=70$

$\left(ii\right)\because ABCDE$ is a part of regular polygon, $CD=CB.\to \left(2\right)$

$\because CDLM$ is a square, $CD=CM.\to \left(3\right)$

$\left(2\right)and\left(3\right)\Rightarrow CB=CM$

$\Rightarrow △BCM$ is an isosceles triangle.

$\Rightarrow \angle CBM=\angle CMB$ in $△BCM$

But $\angle CBM+\angle CMB+\angle BCM={180}^{\circ }$ in $△CBM$ (angle sum property of a triangle)

$\Rightarrow 2\angle CBM+{70}^{\circ }={180}^{\circ }\Rightarrow \angle CBM=\angle CMB={55}^{\circ }$ $\to$ from $\left(1\right)$

By exterior angle property of $△BCM$,

$\angle BCM+\angle CMB=\angle ABM+({180}^{\circ }-{160}^{\circ })\Rightarrow {70}^{\circ }+{55}^{\circ }=y^{\circ }+{20}^{\circ }\Rightarrow y=105$

$\therefore x=70andy=105\left[A\right]$