Question

In the given figure ABCD is rhombus and ALMC is square, AC=BC. Find the $\angle MBC$.

• A. ${60}^0$
• B. ${75}^0$
• C. ${30}^0$
• D. ${45}^0$

Answer 1

Answer: (B)

${75}^0$

In the square ALMC, $\angle ACM={90}^o$

Also, $AC=BC$ (Given)

and $BC=AB$ (Sides of Rhombus)

So, $△ABC$ is an equilateral triangle

Now, $\angle ACB={60}^o$

So from figure , $\angle MCB={90}^o-{60}^o={30}^o$

Now in triangle MBC, BC=CM, so $\angle MBC=\angle BMC=x\left(let\right)$

Sum of all angle of a triangle $={180}^o$

$\angle MCB+\angle MBC+\angle BMC={180}^o$

$\Rightarrow {30}^o+x+x={180}^o$

$\Rightarrow 2x={180}^o-{30}^o$

$\Rightarrow x={75}^0$