a) 30º
b) 45º
c) 60º
d) 90º
Answer: a
Explanation:

Let ∠ACB = θ
$$\frac{{AC}}{{AB}} = \cot \theta $$
$$\frac{{AC}}{{AB}} = \sqrt 3 $$
$$\cot {30^ \circ } = \sqrt 3 $$
θ = 30°
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