Exercise 21 Page 526

For any △ ABC, let the lengths of the sides opposite to angles A, B, and C be a, b, and c, respectively.

sin A/a=sin B/b=sin C/c Let's use this law to find the value of sin ∠ S. Consider the given triangle.

We know that ∠ R is a right angle and the length of the side opposite to it is 5. The lengths of two other sides are TR= 3 and RS= 4. We can use the Law of Sines to find sin ∠ S! sin 90^(∘)/5=sin ∠ S/3 Let's solve the above equation for y using the Cross Product Property.

sin 90^(∘)/5=sin ∠ S/3

CrossMult

Cross multiply

sin 90^(∘)* 3=sin ∠ S* 5

DivEqn

.LHS /5.=.RHS /5.

sin 90^(∘)* 3/5=sin ∠ S

RearrangeEqn

Rearrange equation

sin ∠ S=sin 90^(∘)* 3/5

UseCalc

Use a calculator

sin ∠ S=0.6