C-09-Q204mediumsingle_mcqIn right △ABC\triangle ABC△ABC, hypotenuse AC=2AC = 2AC=2, AB=1AB = 1AB=1. Then ∠ACB=?\angle ACB = ?∠ACB=?a30∘30^\circ30∘b45∘45^\circ45∘cব্যাখ্যাHere ABABAB is opposite ∠ACB\angle ACB∠ACB, so sin(∠ACB)=ABAC=12\sin(\angle ACB)=\dfrac{AB}{AC}=\dfrac{1}{2}sin(∠ACB)=. The angle with this sine is .