The Law of Sines

In this tutorial, we’ll look at the law of sines – one of the important tools that help us solve triangles and find the missing side lengths or angles.

So let’s begin with a quick summary of the convention we use to denote angles and sides in a triangle.

The key is – each side is named by the small case version of the letter for the opposite angle.

Law of sines - labeled triangle

For example, the angle opposite to the side AB is angle C. So its length is represented by c.

The Law of Sines

The law of sines (also known as the sine rule) states that the ratio of side length to the sine of the opposite angle is the same for all sides in a triangle.

So –

asinA=bsinB=bsinC\frac{a}{\sin A} \, \, = \, \, \frac{b}{\sin B} \, \, = \, \, \frac{b}{\sin C}

Simple as that.

Proof of the Law of Sines

Consider the figure below. In the triangle ABC, we drop a perpendicular from A to BC.

Law of sines - proof

Then, taking the sine of angles B and C, we have

sinC=hbbsinC=h(1)\begin{align*} \sin C \, &= \, \frac{h}{b} \\[1.3em] b \, \sin C \, &= \, h \hspace{0.25cm} \rule[0.1cm]{2em}{0.1em} \hspace{0.15cm} (1) \end{align*}
sinB=hccsinB=h(2)\begin{align*} \sin B \, &= \, \frac{h}{c} \\[1.3em] c \, \sin B \, &= \, h \hspace{0.25cm} \rule[0.1cm]{2em}{0.1em} \hspace{0.15cm} (2) \end{align*}

Combining (1)\hspace{0.2em} (1) \hspace{0.2em} and (2)\hspace{0.2em} (2) \hspace{0.2em}, we get

bsinC=csinBbsinB=csinC(3)\begin{align*} b \, \sin C \, &= \, c \, \sin B \\[1em] \frac{b}{\sin B} \, &= \,\frac{c}{\sin C} \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (3) \end{align*}

Next, we drop a perpendicular from B to AC.

Law of sines - proof (a)

And proceeding as above, we have

asinA=csinC(4)\frac{a}{\sin A} \, = \,\frac{c}{\sin C} \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (4)

Combining (3) and (4), we get what we were after – the law of sines.

asinA=bsinB=bsinC\frac{a}{\sin A} \, \, = \, \, \frac{b}{\sin B} \, \, = \, \, \frac{b}{\sin C}


When using the sine rule to find an angle, we need to use the sine inverse function. And the thing with sine inverse is that it gives two possible solutions in the 0o – 180o range. For example,

sinθ=12θ=sin112\begin{align*} \sin \theta \, &= \, \frac{1}{2} \\[1em] \theta \, &= \, \sin^{-1} \frac{1}{2} \end{align*}

Now both sin30°\hspace{0.2em} \sin 30 \degree and sin150°\hspace{0.2em} \sin 150 \degree have a value equal to 12\hspace{0.2em} \frac{1}{2} \hspace{0.2em}. That means,

θ=30°orθ=150°\theta \, = \, 30 \degree \hspace{1em} \text{or} \hspace{1em} \theta \, = \, 150 \degree

So which one to choose? Here are a couple of points to keep in mind.

  • A triangle can have a maximum of one obtuse angle (angle greater than 90o).
  • If there is an obtuse angle, it will be the one opposite the largest side.

It may not be obvious how this will help. Bear with me. It should make more sense when we get to the examples.

Solving Triangles Using the Law of Sines

Now it’s time to see how we can solve triangles using the sine rule.

Note – Often, we need to use the cosine rule along with the sine rule to solve triangles. However, since the focus of this tutorial is on the sine rule, we will only use examples that do not need the cosine rule.

Also, in the following examples, I would encourage you to pay special attention to how the sine rule is being used in the solution and not worry too much about how to solve triangles.

Example 1

In ABC\hspace{0.2em} \triangle ABC \hspace{0.2em}, B=45°\hspace{0.2em} B = 45 \degree \hspace{0.2em}, C=75°\hspace{0.2em} C = 75 \degree \hspace{0.2em}, and BC=12\hspace{0.2em} BC = 12 \hspace{0.2em}. Solve the triangle and get the missing sides and angle.


Step 0.  Before anything else, it’s generally a good idea to draw a rough sketch of the triangle (not necessary, but often makes things easier and helps avoid silly mistakes).

Step 1.  Find the third angle using the angle sum property

A+B+C=180°A+45°+75°=180°A=60°\begin{align*} A + B + C \, &= \, 180 \degree \\[1em] A + 45 \degree + 75 \degree \, &= \, 180 \degree \\[1em] A \, &= \, 60 \degree \end{align*}

Step 2.  Find the missing sides using the sine rule.

So, applying the law of sines,

asinA=bsinB=csinC12sin60°=bsin45°=csin75°\begin{align*} &\frac{a}{\sin A} \, = \, \frac{b}{\sin B} \, = \, \frac{c}{\sin C} \\[1.3em] &\frac{12}{\sin 60 \degree} \, = \, \frac{b}{\sin 45 \degree} \, = \, \frac{c}{\sin 75 \degree} \end{align*}

Now, we split it into two equations and solve for b\hspace{0.2em} b \hspace{0.2em} and c\hspace{0.2em} c \hspace{0.2em}.

12sin60°=bsin45°b=9.80\begin{align*} \frac{12}{\sin 60 \degree} \, &= \, \frac{b}{\sin 45 \degree} \\[1.5em] b \, &= \, 9.80 \end{align*}
12sin60°=csin75°c=13.38\begin{align*} \frac{12}{\sin 60 \degree} \, &= \frac{c}{\sin 75 \degree} \\[1.5em] c \, &= \, 13.38 \end{align*}

That’s it. We have solved the triangle.

Example 2

Solve the triangle below for the missing sides and angles.


Step 1.  Use the Sine Rule to find the missing angle opposite to one of the known sides

Here, we know the sides b\hspace{0.2em} b \hspace{0.2em} and c\hspace{0.2em} c \hspace{0.2em} and the angle BB. So we need to find angle CC.

Applying the sine rule to B\hspace{0.2em} B \hspace{0.2em} and C\hspace{0.2em} C \hspace{0.2em}.

asinA=bsinB7sin80°=5sinBsinB 0.7034B 44.70°\begin{align*} \frac{a}{\sin A} \, &= \, \frac{b}{\sin B} \\[1.5em] \frac{7}{\sin 80 \degree} \, &= \, \frac{5}{\sin B} \\[1.5em] \sin B \, &\approx \ 0.7034 \\[1.5em] B \, &\approx \ 44.70 \degree \end{align*}

Note – We know C\hspace{0.2em} C \hspace{0.2em} is acute because the opposite side, c\hspace{0.2em} c \hspace{0.2em}, isn’t the largest (c<b)\hspace{0.2em} (c < b) \hspace{0.2em}.

Step 2.  Find the third angle using the angle sum property

A+B+C=180°80°+44.70°+C180°C55.3°\begin{align*} A + B + C \, &= \, 180 \degree \\[1em] 80 \degree + 44.70 \degree + C \, &\approx \, 180 \degree \\[1em] C \, &\approx \, 55.3 \degree \end{align*}

Step 3.  Find the missing side using the sine rule

asinA=csinC7sin80°csin55.3°c 5.84\begin{align*} \frac{a}{\sin A} \, &= \, \frac{c}{\sin C} \\[1.5em] \frac{7}{\sin 80 \degree} \, &\approx \, \frac{c}{\sin 55.3 \degree} \\[1.5em] c \, &\approx \ 5.84 \end{align*}


Example 3

In ABC\hspace{0.2em} \triangle ABC \hspace{0.2em}, b=12\hspace{0.2em} b = 12 \hspace{0.2em}, c=16\hspace{0.2em} c = 16 \hspace{0.2em}, and B=40°\hspace{0.2em} B = 40 \degree \hspace{0.2em}. Solve the triangle and get the missing sides and angle.


Step 1.  Use the Sine Rule to find the missing angle opposite to one of the known sides

bsinB=csinC12sin40°=16sinCsinC0.8570\begin{align*} \frac{b}{\sin B} \, &= \, \frac{c}{\sin C} \\[1.5em] \frac{12}{\sin 40 \degree} \, &= \, \frac{16}{\sin C} \\[1.5em] \sin C \, &\approx \, 0.8570 \end{align*}

Now before we take the sine inverse to get the value of C\hspace{0.2em} C \hspace{0.2em}, here's something to keep in mind.

Unlike in the last example, here, we need to consider both acute and obtuse values for C\hspace{0.2em} C. That’s because side c(AB)\hspace{0.2em} c \, \, (AB) \hspace{0.2em} could be the largest side. It is larger than b\hspace{0.2em} b \hspace{0.2em}, and we don’t know a\hspace{0.2em} a \hspace{0.2em} yet.

Acute Solution for B\hspace{0.2em} B \hspace{0.2em}

Csin10.8570°C58.99°\begin{align*} C \, &\approx \, \sin ^{-1} 0.8570 \degree \\[1em] C \, &\approx \, 58.99 \degree \end{align*}

Obtuse Solution for B\hspace{0.2em} B \hspace{0.2em}

C180°58.99°C121.01°\begin{align*} C \, &\approx \, 180 \degree - 58.99 \degree \\[1em] C \, &\approx \, 121.01 \degree \end{align*}

Next, we'll need to work out the remainder of the solution for each of the two cases.

First, let's proceed assuming C\hspace{0.2em} C \hspace{0.2em} is acute.

Step 2.  Find the third angle using the angle sum property

A=180°BC180°40°58.99°81.01°\begin{align*} A \, &= \, 180 \degree - B - C \\[1em] &\approx \, 180 \degree - 40 \degree - 58.99 \degree \\[1em] &\approx \, 81.01 \degree \end{align*}

Step 3.  Find the missing side using the sine rule

asinA=bsinBasin81.01°12sin40°a18.44\begin{align*} \frac{a}{\sin A} \, &= \, \frac{b}{\sin B} \\[1.5em] \frac{a}{\sin 81.01 \degree} \, &\approx \, \frac{12}{\sin 40 \degree} \\[1.5em] a \, &\approx \, 18.44 \end{align*}

Step 4.  Repeat steps 2 and 3 assuming C\hspace{0.2em} C \hspace{0.2em} is obtuse.

So again, first, we find A\hspace{0.2em} A \hspace{0.2em}.

A=180°BC180°40°121.01°18.99°\begin{align*} A \, &= \, 180 \degree - B - C \\[1em] &\approx \, 180 \degree - 40 \degree - 121.01 \degree \\[1em] &\approx \, 18.99 \degree \end{align*}

And then side a\hspace{0.2em} a \hspace{0.2em}

asinA=bsinBasin18.99°12sin40°a6.07\begin{align*} \frac{a}{\sin A} \, &= \, \frac{b}{\sin B} \\[1.5em] \frac{a}{\sin 18.99 \degree} \, &\approx \, \frac{12}{\sin 40 \degree} \\[1.5em] a \, &\approx \, 6.07 \end{align*}

And that brings us to the end of this tutorial on the law of sines. Until next time.