In ΔABC, ∠B = 60° and ∠C = 40°. If AD and AE be respectively the internal bisector of ∠A and perpendicular on BC, then the measure of ∠DAE is

According to question,

Given :
∠B = 60°
∠C = 40°
As we know that
∠A + ∠B + ∠C = 180°
∠A = 180° - 60° - 40°
∠A = 80°
∴ ∠BAD = $$\frac{{{{80}^ \circ }}}{2}$$ = 40°
In ΔAEB
∠A + ∠B + ∠E = 180°
∠A = 180° - 60° - 90°
∠A = 30°
Then,
∠DAE = ∠DAB - ∠EAB
∠DAE = 40° - 30°
∠DAE = 10°

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By Trick
$$\eqalign{ & \angle DAE = \frac{{\angle B - \angle C}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{60}^ \circ } - {{40}^ \circ }}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {10^ \circ } \cr} $$