Exercise 13 Page 675

For any △ ABC, let the lengths of the sides opposite 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 x. Consider the given triangle.

We know that the length of a side is 5 and that the measure of its opposite angle is 68. We want to find the length of the side that is opposite to the angle whose measure is 50. We can use the Law of Sines again! sin 68/5=sin 50/x Let's solve the above equation for x using the Cross Product Property.

sin 68/5=sin 50/x

▼

Solve for x

CrossMult

Cross multiply

sin 68* x=sin 50* 5

DivEqn

.LHS /sin 68.=.RHS /sin 68.

x=sin 50* 5/sin 68

UseCalc

Use a calculator

x=4.131027...

RoundDec

Round to 1 decimal place(s)

x≈ 4.1