If \(\sec A = 13/5\) and A is acute, find \(\sin A\).
12/13
\(\sec A = \tfrac{1}{\cos A} = \tfrac{13}{5}\), so \(\cos A = \tfrac{5}{13}\).
In a right triangle with acute angle A: adjacent = 5, hypotenuse = 13.
By the Pythagoras theorem: opposite = \(\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\).
\(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{12}{13}\).
Hence, the answer is 12/13.
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