Question

∠A is an acute angle in a right triangle.
Given that cosA=12\13, what is the ratio for sinA?
Enter your answer in the boxes as a fraction in simplest form.

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Answer 1

5/ 13 is your answer

Answer 2

Cos A = 12/13

$sin^{2}A$ = $1 - cos^{2}A$

Plug in the value of cos A = 12/13

$sin^{2}A$ = $1 - \left ( \frac{12}{13} \right )^{2}$

$sin^{2}A$ = $1 - \left ( \frac{144}{169} \right )$

$sin^{2}A$ = $\frac{169-144}{169}$

$sin^{2}A$ = $\frac{25}{169}$

Sin A = $\pm$ (5/13)