Question : In $\triangle A B C, \mathrm{BD} \perp \mathrm{AC}$ at $\mathrm{D}$. $\mathrm{E}$ is a point on $\mathrm{BC}$ such that $\angle B E A=x^{\circ}$. If $\angle E A C=46^{\circ}$ and $\angle E B D=60^{\circ}$, then the value of $x$ is:
Option 1: 72°
Option 2: 78°
Option 3: 68°
Option 4: 76°
Correct Answer: 76°
Solution :
$\angle B E A=x^{\circ}$
$\angle E A C=46^{\circ}$
$\angle E B D=60^{\circ}$
In $\triangle BDC$,
$\angle DCB + \angle BDC + \angle CBD = 180^{\circ}$
⇒ $\angle DCB +90^{\circ}+60^{\circ} = 180^{\circ}$
⇒ $\angle DCB = 180^{\circ} - 90^{\circ}-60^{\circ}$
⇒ $\angle DCB = 30^{\circ}$
In $\triangle AEC$,
$\angle AEC + \angle ECA + \angle CAE = 180^{\circ}$
⇒ $\angle AEC +30^{\circ}+46^{\circ} = 180^{\circ}$
⇒ $\angle AEC = 180^{\circ} - 46^{\circ}-30^{\circ}$
⇒ $\angle AEC = 104^{\circ}$
$\angle B E A+ \angle AEC = 180^{\circ}$
⇒ $ x^{\circ} + 104^{\circ} = 180^{\circ}$
⇒ $ x^{\circ} = 180^{\circ} - 104^{\circ} = 76^{\circ}$
Hence, the correct answer is $76^{\circ}$.
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