Question

Given $ABCD$ is a rhombus with $\angle DAB=2x+{15}^{\circ },\angle DCB=3x-{30}^{\circ },\angle BDC$ equals

• A. ${45}^{\circ }$
• B. ${35}^{\circ }$
• C. ${37.5}^{\circ }$
• D. ${42.5}^{\circ }$

Answer 1

Answer: (C)

${37.5}^{\circ }$

Here, $ABCD$ is a rhombus.

$\angle DAB=2x+{15}^o$ and $\angle DCB=3x-{30}^o$

We know that opposite angles of a rhombus are equal.

$\therefore$ $\angle DAB=\angle DCB$

$\Rightarrow$ $2x+{15}^o=3x-{30}^o$

$\Rightarrow$ $x={45}^o$

$\Rightarrow$ $\angle DAB=2\times {45}^o+{15}^o={105}^o$

$\therefore$ $\angle DAB=\angle DCB={105}^o$

Sum of four angles of a rhombus is ${360}^o$

$\therefore$ $\angle DAB+\angle ABC+\angle DCB+\angle CDA={360}^o$

$\Rightarrow$ ${105}^o+\angle ABC+{105}^o+\angle CDA={360}^o$

$\Rightarrow$ ${210}^o+\angle CDA+\angle CDA={360}^o$

$\Rightarrow$ $2\angle CDA={150}^o$

$\Rightarrow$ $\angle CDA={75}^o$

Here, $BD$ is a diagonal of rhombus which bisect $\angle CDA$

$\therefore$ $\angle BDC=\frac{1}{2}\angle CDA=\frac{1}{2}\times {75}^o={37.5}^o$

$\therefore$ $\angle BDC={37.5}^o$