Question

How do I solve for what is sin A for the triangle shown

Answer 1

Option a:

sin A for the triangle is $\frac{3}{5}$.

Solution:

The given triangle is a right triangle.

The adjacent side to angle A is AC.

The opposite side to angle A is BC.

Hypotenuse is AB.

AC = 8, BC = 6 and AB = 10.

Using trigonometric ratio formulas,

$\sin \theta=\frac{\text { Opposite side }}{\text { Hypotenuse }}$

$\sin A=\frac{BC}{AB}$

$\sin A=\frac{6}{10}$

Divide both numerator and denominator by 2, we get

$\sin A=\frac{6\div2}{10\div 2}$

$\sin A=\frac{3}{5}$

Hence sin A for the triangle is $\frac{3}{5}$.

Option a is the correct answer.