Question : In $\triangle \mathrm{ABC}, \mathrm{BD} \perp \mathrm{AC}$ at D. E is a point on BC such that $\angle \mathrm{BEA}=x^{\circ}$. If $\angle \mathrm{EAC}=62^{\circ}$ and $\angle \mathrm{EBD}=60^{\circ}$, then the value of $x$ is:
Option 1: $92^\circ$
Option 2: $78^\circ$
Option 3: $76^\circ$
Option 4: $68^\circ$
Correct Answer: $92^\circ$
Solution :
In $\triangle AOD$,
$\angle AOD + \angle ADO + \angle DAO = 180^\circ$
⇒ $\angle AOD + 90^\circ + 62^\circ = 180^\circ$
⇒ $\angle AOD = 180^\circ - 152^\circ$
⇒ $\angle AOD = 28^\circ$
$\because \angle AOD$ and $\angle EOB$ are vertical opposite angles,
$\angle AOD = \angle EOB = 28^\circ$
In $\triangle BOE$,
$\angle BEO + \angle EOB + \angle OBE = 180^\circ$
⇒ $x + 28^\circ + 60^\circ = 180^\circ$
⇒ $x = 180^\circ - 88^\circ$
⇒ $x = 92^\circ$
Hence, the correct answer is $92^\circ$.
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