In ΔABC, ∠BAC = 90° and AB = 1/2 BC, Then the measure of ∠ACB is :

Answer: Option B

According to question,
Given :
BAC is right angle triangle
AB = $$\frac{1}{2}$$BC

$$\begin{array}{rl}& \frac{AB}{BC}=\frac{1}{2}\frac{P}{H}=\frac{1}{2}\\ & \sin \theta =\frac{P}{H}=\frac{1}{2}\\ & \sin 30^{\circ }=\frac{1}{2}\\ & \therefore \theta =\angle ACB=30^{\circ }\end{array}$$