Question

Right triangle ABC has its right angle at C.

Answer 1

The answers are:

$cosA=\frac{12}{13}$

$sinB=\frac{12}{13}$

$sinA=\frac{5}{13}$

The explanation is shown below:

1. By definition, when you have a rigth triangle, you can find the sine, cosine and tangent as following:

$sine=\frac{opposite}{hypotenuse}\\cosine=\frac{adjacent}{hypotenuse}\\tangent=\frac{opposite}{adjacent}$

3. To know the lenght of the hypotenuse you can apply the Pythagorean Theorem:

$hypotenuse=\sqrt{12^{2}+5^{2}}=13$

4. Therefore if AC=12, BC=5 and the right angle is at C, you can substitute values to find the sine, cosine and tangent of A and B:

$sinA=\frac{5}{13}$

$sinB=\frac{12}{13}$

$cosA=\frac{12}{13}$

$cosB=\frac{5}{13}$

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

$tanB=\frac{12}{5}$