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 $ an \left ( \angle ACB ight )=6 an \left ( \angle DBE ight )$, then $\angle ACB =$
Option 1: $30^{\circ}$
Option 2: $45^{\circ}$
Option 3: $60^{\circ}$
Option 4: $15^{\circ}$

Question

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 $ an \left ( \angle ACB ight )=6 an \left ( \angle DBE ight )$, then $\angle ACB =$

Option 1: $30^{\circ}$

Option 2: $45^{\circ}$

Option 3: $60^{\circ}$

Option 4: $15^{\circ}$

Team Careers360 · Accepted Answer

Correct Answer: $60^{\circ}$

Solution :

Given that $AE:ED=5:1$, $\angle BAD=30^{\circ}$ and $ an \left ( \angle ACB ight )=6 an \left ( \angle DBE ight )$
Let $AE$ and $ED$ be $5x,x$.
So, $DE=x, AE=5x$ and $AD=6x$
In $ riangle ABD$,
$\angle ABD+\angle BAD=90^{\circ}$
⇒ $\angle ABD =90^{\circ}-\angle BAD$
⇒ $\angle ABD =90^{\circ}-30^{\circ}=60^{\circ}$
⇒ $ an \left ( \angle ACD ight )=\frac{AD}{DC}=\frac{6x}{DC}$
and, $ an \left ( \angle DBE ight )=\frac{DE}{BD}=\frac{x}{BD}$
From the given expression,
$ an \left ( \angle ACB ight )=6 an \left ( \angle DBE ight )$
⇒ $\frac{6x}{DC}=\frac{6x}{BD}$
⇒ $BD=DC$
⇒ $ riangle ABD \cong riangle ACD$
Since $\angle ABD=60^{\circ}$ then, $\angle ACB=60^{\circ}$
Hence, the correct answer is $60^{\circ}$.

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Question : D is a point on the side BC of a triangle ABC such that AD⊥BC. E is a point on AD for which AE:ED=5:1. If ∠BAD=30∘ and tan⁡(∠ACB)=6tan⁡(∠DBE), then ∠ACB=Option 1: 30∘Option 2: 45∘Option 3: 60∘Option 4: 15∘

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