The law of sines

In a triangle, the ratio of a side to the sine of the angle opposite it is the same for all three sides, and it is the diameter of the circle that circumscribes the triangle:

$\displaystyle a$ opposite $\displaystyle \alpha$, $\displaystyle b$ opposite $\displaystyle \beta$, $\displaystyle c$ opposite $\displaystyle \gamma$; $\displaystyle R$ – the radius of the circumscribed circle. Opposite the larger side – the larger sine.

The proof below draws the height $\displaystyle {CD}={h}$ to the side $\displaystyle AB$, and works it out in two ways.

• In the right-angled triangle $\displaystyle ADC$: $\displaystyle {h}={b\sin\alpha}$.
• In the right-angled triangle $\displaystyle BDC$: $\displaystyle {h}={a\sin\beta}$.
• So $\displaystyle {b\sin\alpha}={a\sin\beta}$, and dividing by $\displaystyle \sin\alpha\sin\beta$: $\displaystyle {\frac{a}{\sin\alpha}}={\frac{b}{\sin\beta}}$.

A height from another vertex gives $\displaystyle \frac{c}{\sin\gamma}$ the same way. When one of the angles is obtuse, use $\displaystyle {\sin(180^{\circ}-\alpha)}={\sin\alpha}$, as in the proof of the area formula.

The proof below draws the diameter $\displaystyle BD$ of the circumscribed circle, and shows that $\displaystyle {a}={2R\sin\alpha}$.

• $\displaystyle {\angle BDC}={\angle BAC}={\alpha}$: inscribed angles on the same arc $\displaystyle BC$.
• $\displaystyle {\angle BCD}={90^{\circ}}$: an inscribed angle on a diameter.
• In the right-angled triangle $\displaystyle BCD$ the hypotenuse is $\displaystyle {BD}={2R}$, and $\displaystyle {BC}={a}$ is the leg opposite $\displaystyle \alpha$: $\displaystyle {a}={2R\sin\alpha}$.

So $\displaystyle {\frac{a}{\sin\alpha}}={2R}$, and the same way for the other two sides. (When $\displaystyle \alpha$ is obtuse, $\displaystyle D$ and $\displaystyle A$ are on different sides of $\displaystyle BC$, and the angle at $\displaystyle D$ is $\displaystyle 180^{\circ}-\alpha$ – with the same sine.)

When two angles and a side are known, the third angle is $\displaystyle 180^{\circ}$ minus both, and the law of sines gives every side.

The example below has $\displaystyle {a}={10}$, $\displaystyle {\beta}={50^{\circ}}$, $\displaystyle {\gamma}={60^{\circ}}$, and finds $\displaystyle b$:

• $\displaystyle {\alpha}={180^{\circ}-50^{\circ}-60^{\circ}}={70^{\circ}}$;
• $\displaystyle {\frac{b}{\sin50^{\circ}}}={\frac{10}{\sin70^{\circ}}}$, so $\displaystyle {b}={\frac{10\sin50^{\circ}}{\sin70^{\circ}}\approx8.15}$.

When two sides and the angle opposite one of them are known, the law of sines gives the sine of the angle opposite the other. A sine has two angles between $\displaystyle 0^{\circ}$ and $\displaystyle 180^{\circ}$: $\displaystyle \beta$ and $\displaystyle 180^{\circ}-\beta$.

• The given angle opposite the larger side – only the acute $\displaystyle \beta$ fits (the angle opposite the smaller side is smaller), and there is one triangle. This is the fourth congruence theorem: two sides and the angle opposite the larger of them fix the triangle.
• The given angle opposite the smaller side – sometimes both angles fit, and there are two different triangles with the same data.

In every case check that the angles add up to less than $\displaystyle 180^{\circ}$.

From the law of sines, $\displaystyle {b}={\frac{a\sin\beta}{\sin\alpha}}$, and $\displaystyle {\sin\alpha}={\sin(180^{\circ}-\beta-\gamma)}={\sin(\beta+\gamma)}$. Substitute in $\displaystyle {S}={\frac{1}{2}ab\sin\gamma}$:

$\displaystyle \beta$ and $\displaystyle \gamma$ are the angles at the side $\displaystyle a$. In the previous example: $\displaystyle {S}={\frac{100\sin50^{\circ}\sin60^{\circ}}{2\sin110^{\circ}}\approx35.30}$.

Work out the exact value.

In a triangle: $\displaystyle {a}={10}$, $\displaystyle {\alpha}={30^{\circ}}$, $\displaystyle {\beta}={45^{\circ}}$. Work out $\displaystyle b$.

In a triangle: $\displaystyle {a}={6}$, $\displaystyle {\alpha}={60^{\circ}}$. Work out the radius $\displaystyle R$ of the circumscribed circle.

In triangle $\displaystyle ABC$: $\displaystyle {BC}={4}$, $\displaystyle {AC}={4\sqrt{2}}$, and $\displaystyle {\angle A}={30^{\circ}}$.
$\displaystyle x$ is the angle $\displaystyle B$. Find all the possibilities.

Two sides of a triangle are $\displaystyle 7$ cm and $\displaystyle 5$ cm.
The angle opposite the side of $\displaystyle 5$ cm is $\displaystyle 40^{\circ}$.
Round to two decimal places.

Work out the acute angle opposite the side of $\displaystyle 7$ cm.

Work out the other possibility for the same angle.

How many triangles fit the data? Write the angles of each.

Two sides of a triangle are $\displaystyle 5$ cm and $\displaystyle 7$ cm, and the angle opposite the side of $\displaystyle 7$ cm is $\displaystyle 40^{\circ}$.
Work out the angle opposite the side of $\displaystyle 5$ cm, rounded to two decimal places. Why is there only one possibility here?