Question

Calculate the ratio of the length of the longer diagonal to the length of the shorter diagonal of a rhombus if, its one angle measures ${60}^o$.

• A. $2$
• B. $\sqrt{3}$
• C. $\sqrt{2}$
• D. $\frac{\sqrt{3}}{2}$
• E. $\frac{\sqrt{2}}{2}$

Answer 1

The correct option is B $\sqrt{3}$

Let ABCD be a rhombus of each side 'a' let $\angle ADC={60}^0$

We know that diagonals of rhombus bisect each other at ${90}^0$.

In $△AOB,\angle ADO={30}^0,\angle AOD={90}^0$ and $\angle DAO={60}^0$

Now, $\sin {30}^0=\frac{AO}{AD}\Rightarrow AO=AD\sin {30}^0=a\times \frac{1}{2}=\frac{a}{2}$

$\cos {30}^0=\frac{DO}{AD}\Rightarrow DO=AD\cos {30}^0=a\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}a}{2}$

$\therefore AC=2AO=2\times \frac{a}{2}=a$ & $BD=2DO=2\times \frac{\sqrt{3}a}{2}=\sqrt{3}a$

$\therefore \frac{BD}{AC}=\frac{\sqrt{3}a}{a}=\sqrt{3}$