Question

5sinA +12 cosA =13 find tanA

Answer 1

$\textbf{Given:}$

$5\;sinA+12\;cosA=13$

$\textbf{To find: tanA}$

$5\;sinA+12\;cosA=13$

$\text{Squaring on bothsides we get}$

$(5\;sinA+12\;cosA)^2=169$

$25\;sin^2A+144\;cos^2A+120\;sinA\;cosA=169$

$25(1-cos^2A)+144(1-sin^2A)+120\;sinA\;cosA=169$

$25-25\;cos^2A+144-144\;sin^2A)+120\;sinA\;cosA=169$

$-25\;cos^2A-144\;sin^2A+120\;sinA\;cosA=0$

$25\;cos^2A+144\;sin^2A-120\;sinA\;cosA=0$

$(12\;sinA-5\;cosA)^2=0$

$12\;sinA-5\;cosA=0$

$\text{Now, we have two equations}$

$5\;sinA+12\;cosA-13=0$

$12\;sinA-5\;cosA-0=0$

$\text{We apply cross multiplication rule to solve the equations}$

$\displaystyle\frac{sinA}{0-65}=\frac{cosA}{-156-0}=\frac{1}{-25-144}$

$\displaystyle\frac{sinA}{-65}=\frac{cosA}{-156}=\frac{1}{-169}$

$\implies\displaystyle\;sinA=\frac{-65}{-169}=\frac{5}{13}$ $\;\text{and}$

$\implies\displaystyle\;cosA=\frac{-156}{-169}=\frac{12}{13}$

$tanA=\displaystyle\frac{sinA}{cosA}$

$tanA=\displaystyle\frac{\frac{5}{13}}{\frac{12}{13}}$

$\implies\boxed{\bf\,tanA=\displaystyle\frac{5}{12}}$

$\textbf{Find more:}$

1.If 6 tan A -5 = 0 find the value of( 3 Sin A - Cos A) / (5 Cos A + 9 sin A)

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2.CotA+cosecA=3,then find sinA=?​

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