C-09-Q191mediumsingle_mcqIf cos(A−B)=1\cos(A - B) = 1cos(A−B)=1 and 2sin(A+B)=32\sin(A + B) = \sqrt{3}2sin(A+B)=3 with A,BA, BA,B acute, then A=?A = ?A=?a30∘30^\circ30∘b45∘45^\circ45∘c60∘60^\circ60∘d15∘15^\circ15∘ব্যাখ্যাSince cos(A−B)=1\cos(A-B)=1cos(A−B)=1, A−B=0∘A-B=0^\circA−B=0∘, so A=BA=BA=B. Then 2sin(A+B)=32\sin(A+B)=\sqrt32sin(A+B)=3 gives sin2A=32\sin 2A=\tfrac{\sqrt3}{2}sin2A=23, so 2A=60∘2A=60^\circ2A=60∘ and A=30∘A=30^\circA=30∘.