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In ΔABC, ∠BAC = 90° and AB = $$\frac{1}{2}$$ BC, Then the measure of ∠ACB is :
Answer & Solution
Correct Answer:
Option
B
According to question,
Given :
BAC is right angle triangle
AB = $$\frac{1}{2}$$BC
$$\eqalign{ & \frac{{AB}}{{BC}} = \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{P}{H} = \frac{1}{2} \cr & \sin \theta = \frac{P}{H} = \frac{1}{2} \cr & \sin \,{30^ \circ } = \frac{1}{2} \cr & \therefore \theta = \angle ACB = {30^ \circ } \cr} $$
Given :
BAC is right angle triangle
AB = $$\frac{1}{2}$$BC
$$\eqalign{ & \frac{{AB}}{{BC}} = \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{P}{H} = \frac{1}{2} \cr & \sin \theta = \frac{P}{H} = \frac{1}{2} \cr & \sin \,{30^ \circ } = \frac{1}{2} \cr & \therefore \theta = \angle ACB = {30^ \circ } \cr} $$
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