Question

In the given figure, if ABCD is a rhombus with $\angle ADC=58°$, then the measure of 2$\angle CAB–\angle ABD$ is equal to

• A. ${30}^0$
• B. ${25}^0$
• C. ${38}^0$
• D. ${93}^0$

Answer 1

The correct option is D ${93}^0$


We know that diagonals of a rhombus bisect their vertex angles.
Also, the opposite angles of a rhombus are equal.
$\therefore \angle ABC=\angle ADC=58°$
And, $\angle ABD=\frac{1}{2}\times {58}^0={29}^0$
Since the sum of angles of a quadrilateral is $360°$.
$\therefore \angle DAB+\angle ABC+\angle BCD+\angle CDA=360°$
$\Rightarrow \angle DAB+58°+\angle BCD+58°=360°$
$\Rightarrow 2\angle DAB=360°–116°=244°$ ($\because \angle DAB=\angle BCD$)
$\Rightarrow \angle DAB=122°$
Also $\angle CAB=\frac{1}{2}\times {122}^0={61}^0$
$\therefore 2\angle CAB-\angle ABD=2\left({61}^0\right)-{29}^0$
$={122}^0-{29}^0$
$={93}^0$
Hence, the correct answer is option (d).