# How does the Law of Sines work?

My teacher gave me the formula for the Law of Sines and I know how to solve questions like this, but I don't see how the theorem below can actually work.

Can someone please explain it to me?

• Are you asking for a proof of the Law of Sines? – Ted Shifrin May 28 '14 at 0:16
• I don't understand how this law works. – IHeartBunnies May 28 '14 at 0:20
• Do you understand how the double equal signs work? – Brad May 28 '14 at 0:24
• What do you mean? – IHeartBunnies May 28 '14 at 0:40

Let's look at the right triangle below. I use a right triangle so that we can illustrate the law of sines by applying the familiar SOH-CAH-TOA.

As an example, let's look at angles $A$ and $C$.

• We see that $\displaystyle\sin(A)=\frac{a}{b}=\frac{\text{opposite side of A}}{\text{hypotenuse}}$. This gives $\displaystyle \frac{1}{b}=\frac{\sin(A)}{a}$
• We see that $\displaystyle\sin(C)=\frac{c}{b}=\frac{\text{opposite side of C}}{\text{hypotenuse}}$. This gives $\displaystyle \frac{1}{b}=\frac{\sin(C)}{c}$

Since we have two expressions for $\displaystyle\frac{1}{b}$, we can set the expressions equal to each other and obtain $$\frac{\sin(A)}{a}=\frac{\sin(C)}{c}$$ This is the kind of relation that is fundamental in the Law of Sines.

It turns out that this relation holds for any two angles that you pick (so not restricted to angles $A$ or $C$). It also turns out that the Law of Sines holds for not just right angles, but also acute and obtuse angles as well.

However, the approach for deriving the Law of Sines for acute and obtuse are different; I only showed the approach for right angles. But please ask further if you'd like to see more explanation of how this Law of Sines works for acute/obtuse angles.

Consider the hight dropped from the vertex $B$, giving two right triangles. Then $\sin(A)=\frac{h}{c}$ and $\sin(C)=\frac{h}{a}$ or $c\sin(A)=a\sin(C)$ this gives one of the three equalities, the others being the same.