Question : In $\triangle ABC, \angle B = 60^\circ$ and $\angle C = 40^\circ$, AD and AE are respectively the bisectors of $\angle A$ and perpendicular on BC. Find the measure of $\angle EAD$.
Option 1: $11^\circ$
Option 2: $10^\circ$
Option 3: $12^\circ$
Option 4: $9^\circ$
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Correct Answer: $10^\circ$
Solution :
Given: In $\triangle ABC$, $\angle B = 60^\circ$ and $\angle C = 40^\circ$.
We know that,
$\angle BAC+\angle ABC+\angle ACB=180^\circ$
⇒ $\angle BAC = 180^\circ-60^\circ-40^\circ$
⇒ $\angle BAC = 80^\circ$
Since AD is the angle bisector of $\angle BAC$ in $\triangle ABC$
$\therefore \angle BAD = \frac{80^\circ}{2}=40^\circ$
Now, in $\triangle AEB$
$\angle BAE+\angle ABE+\angle AEB=180^\circ$
⇒ $\angle BAE = 180^\circ-60^\circ-90^\circ$
⇒ $\angle BAE = 30^\circ$
⇒ $\angle EAB = 30^\circ$
$\therefore \angle EAD=\angle BAD-\angle EAB=40^\circ-30^\circ=10^\circ$
Hence, the correct answer is $10^\circ$.
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