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Given (A) = 5/13 and (B) = 3/5, where A and B are acute angles, find (A+B). Also, given (A) = 1/3 and (B) = 1/4, find (A+B).

Mathematics

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1. Given sin⁡(A)=5/13\sin(A) = 5/13 and cos⁡(B)=3/5\cos(B) = 3/5, where A and B are acute angles, find sin⁡(A+B)\sin(A+B).

  1. Find cos⁡(A)\cos(A) and sin⁡(B)\sin(B).

    Since A is acute, cos⁡(A)=1−sin⁡2(A)=1−(513)2=1−25169=144169=1213\cos(A) = \sqrt{1-\sin^2(A)} = \sqrt{1-\left(\frac{5}{13}\right)^2} = \sqrt{1-\frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}.
    Since B is acute, sin⁡(B)=1−cos⁡2(B)=1−(35)2=1−925=1625=45\sin(B) = \sqrt{1-\cos^2(B)} = \sqrt{1-\left(\frac{3}{5}\right)^2} = \sqrt{1-\frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}.

  2. Apply the sum formula for sine.

    sin⁡(A+B)=sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B)\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B) =(513)(35)+(1213)(45)= \left(\frac{5}{13}\right)\left(\frac{3}{5}\right) + \left(\frac{12}{13}\right)\left(\frac{4}{5}\right) =1565+4865= \frac{15}{65} + \frac{48}{65} =15+4865= \frac{15+48}{65} =6365= \frac{63}{65}

    The final answer is 6365\boxed{\frac{63}{65}}.

    2. Given tan⁡(A)=1/3\tan(A) = 1/3 and tan⁡(B)=1/4\tan(B) = 1/4, find tan⁡(A+B)\tan(A+B).

  3. Apply the sum formula for tangent.

    tan⁡(A+B)=tan⁡(A)+tan⁡(B)1−tan⁡(A)tan⁡(B)\tan(A+B) = \frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} =13+141−(13)(14)= \frac{\frac{1}{3}+\frac{1}{4}}{1-\left(\frac{1}{3}\right)\left(\frac{1}{4}\right)} =412+3121−112= \frac{\frac{4}{12}+\frac{3}{12}}{1-\frac{1}{12}} =7121212−112= \frac{\frac{7}{12}}{\frac{12}{12}-\frac{1}{12}} =7121112= \frac{\frac{7}{12}}{\frac{11}{12}} =711= \frac{7}{11}

    The final answer is 711\boxed{\frac{7}{11}}.

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