Week 16 – Precalculus 11 – Sine Law

Hello! This week I will be covering the sine law in the trigonometry unit. The sin law is a very useful tool for solving triangles.

The sine law works for every triangle and the equation for it is: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ .

This means that when side a is divided by the sine of angle A, it is equal to the side b divided by the sine of angle B, which is equal to the side c divided by the sine of angle C.

triangle 35 degrees, 105 degrees, 7

Say you had to calculate side c.

Recall the law of sines. $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ . First put in the values that we know: $\frac{a}{\sin A}=\frac{7}{\sin35}=\frac{c}{\sin105}$ . Ignore $\frac{a}{\sin A}$ as it does not have values that are useful to us.

Cross multiply to rearrange the equation and solve for side c.

We should have the answer of c = 11.8.

triangle 63 degrees, 4.7, 5.5

Say you had to calculate angle B.

Rearrange the law of sines to get $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ . First put in the values that we know: $\frac{\sin A}{a}=\frac{\sin B}{4.7}=\frac{\sin63}{5.5}$ . Ignore $\frac{\sin A}{a}$ as it does not have values that are useful to us.

Like before, cross multiply to rearrange the equation and solve for angle C.

Use inverse sine to get rid of sin B = 0.7614… and isolate the variable B.

Our answer should be B = 49.6.

Hope this helps!

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