Question : $D$ is a point on the side $BC$ of a triangle $ABC$ such that $AD\perp BC$. $E$ is a point on $AD$ for which $AE:ED=5:1$. If $\angle BAD=30^{\circ}$ and $\tan \left ( \angle ACB \right )=6\tan \left ( \angle DBE \right )$, then $\angle ACB =$
Option 1: $30^{\circ}$
Option 2: $45^{\circ}$
Option 3: $60^{\circ}$
Option 4: $15^{\circ}$
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Correct Answer: $60^{\circ}$
Solution :
Given that $AE:ED=5:1$, $\angle BAD=30^{\circ}$ and $\tan \left ( \angle ACB \right )=6\tan \left ( \angle DBE \right )$
Let $AE$ and $ED$ be $5x,x$.
So, $DE=x, AE=5x$ and $AD=6x$
In $\triangle ABD$,
$\angle ABD+\angle BAD=90^{\circ}$
⇒ $\angle ABD =90^{\circ}-\angle BAD$
⇒ $\angle ABD =90^{\circ}-30^{\circ}=60^{\circ}$
⇒ $\tan \left ( \angle ACD \right )=\frac{AD}{DC}=\frac{6x}{DC}$
and, $\tan \left ( \angle DBE \right )=\frac{DE}{BD}=\frac{x}{BD}$
From the given expression,
$\tan \left ( \angle ACB \right )=6\tan \left ( \angle DBE \right )$
⇒ $\frac{6x}{DC}=\frac{6x}{BD}$
⇒ $BD=DC$
⇒ $\triangle ABD \cong \triangle ACD$
Since $\angle ABD=60^{\circ}$ then, $\angle ACB=60^{\circ}$
Hence, the correct answer is $60^{\circ}$.
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