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Introduction to Trigonometry · Class 10 · CBSE Mathematics

If sin(A - B) = frac{1}{2} and cos(A + B) = frac{1}{2}, where (A + B) is acute and A > B, find the values of A and B.

Practice question from Introduction to Trigonometry (Class 10 · CBSE Mathematics), with the correct answer and a step-by-step solution.

medium · Subjective · Q-BFF296

If sin⁡(A−B)=12\sin(A - B) = \frac{1}{2} and cos⁡(A+B)=12\cos(A + B) = \frac{1}{2}, where (A+B)(A + B) is acute and A>BA > B, find the values of AA and BB.

Answer: Since sin⁡(A−B)=12\sin(A - B) = \frac{1}{2}, we know that A−B=30∘A - B = 30^\circ. Similarly, since cos⁡(A+B)=12\cos(A + B) = \frac{1}{2}, we have A+B=60∘A + B = 60^\circ. Adding both equations gives 2A=90∘2A = 90^\circ, so A=45∘A = 45^\circ. Substituting AA back gives B=15∘B = 15^\circ.

Solution

Given sin⁡(A−B)=sin⁡30∘  ⟹  A−B=30∘\sin(A - B) = \sin 30^\circ \implies A - B = 30^\circ. Also, cos⁡(A+B)=cos⁡60∘  ⟹  A+B=60∘\cos(A + B) = \cos 60^\circ \implies A + B = 60^\circ. Solving the simultaneous linear equations, we get 2A=90∘  ⟹  A=45∘2A = 90^\circ \implies A = 45^\circ and B=15∘B = 15^\circ.

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