Area of a Triangle Using Sine

You already know that the area of a triangle is half the base times the height. But in most real problems, like a triangular plot of land, nobody hands you the height. If you know two sides and the angle between them, trigonometry gives you the height for free, and that leads to one of the most useful formulas in the course.

Key ideas

The formula

For any triangle $ABC$, with sides $a$, $b$, $c$ opposite the angles $A$, $B$, $C$:

$$\text{Area} = \frac{1}{2}ab\sin C$$

In words: half the product of two sides, times the sine of the angle between them (the included angle). This formula is in the formula booklet. Because you can choose any two sides, it can also be written $\dfrac{1}{2}bc\sin A$ or $\dfrac{1}{2}ac\sin B$.

Where it comes from

Take $a = CB$ as the base and drop a perpendicular from $A$ to it. Call its length $h$.

The height is the side opposite $C$ in a right triangle with hypotenuse $b$, so $h = b\sin C$.

In the right triangle on the left, $\sin C = \dfrac{h}{b}$, so $h = b\sin C$. Then

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times a \times b\sin C = \frac{1}{2}ab\sin C$$

If $C$ is obtuse, the height falls outside the triangle, and the height is $b\sin(180^\circ - C)$. Since $\sin(180^\circ - C) = \sin C$, the same formula still works.

Which tool do you need?

The area formula needs two sides and the included angle. If you don’t have that, get it first:

You know	First step	Then
Two sides and the included angle (SAS)	nothing	use $\frac{1}{2}ab\sin C$
Three sides (SSS)	cosine rule to find an angle	use $\frac{1}{2}ab\sin C$
Two angles and a side (AAS or ASA)	angle sum, then sine rule for a second side	use $\frac{1}{2}ab\sin C$

Finding an angle from the area

If you know the area and two sides, rearrange:

$$\sin C = \frac{2 \times \text{Area}}{ab}$$

Since $\sin C = \sin(180^\circ - C)$, there are usually two possible angles: an acute one, $C$, and an obtuse one, $180^\circ - C$. Both give a triangle with the same area. The question will often tell you which one it wants (“the angle is obtuse”); if not, give both.

Worked examples

Example 1: Two sides and the included angle

In triangle $ABC$, $a = 7$ cm, $b = 9$ cm and $C = 40^\circ$. Find the area.

Solution. The angle $C$ is between sides $a$ and $b$, so use the formula directly:

$$\text{Area} = \frac{1}{2}(7)(9)\sin 40^\circ = 20.247\ldots \approx 20.2 \text{ cm}^2 \quad (\text{3 s.f.})$$

Example 2: Three sides

A triangle has sides $5$ cm, $7$ cm and $8$ cm. Find its area, exactly.

Solution. First find an angle with the cosine rule. Let $B$ be the angle opposite the $7$ cm side, so it sits between the sides $5$ and $8$:

$$\cos B = \frac{5^2 + 8^2 - 7^2}{2(5)(8)} = \frac{40}{80} = \frac{1}{2} \quad\Rightarrow\quad B = 60^\circ$$

Now use the two sides that enclose $B$:

$$\text{Area} = \frac{1}{2}(5)(8)\sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.3 \text{ cm}^2$$

Example 3: Finding an angle from the area

Triangle $PQR$ has $PQ = 10$ cm, $PR = 12$ cm, and area $45 \text{ cm}^2$. Find the two possible sizes of angle $P$, and the length of $QR$ in each case.

Solution. Angle $P$ is between the sides $10$ and $12$:

$$\frac{1}{2}(10)(12)\sin P = 45 \quad\Rightarrow\quad \sin P = \frac{45}{60} = 0.75$$ $$P = 48.590\ldots^\circ \approx 48.6^\circ \qquad\text{or}\qquad P = 180^\circ - 48.590\ldots^\circ = 131.409\ldots^\circ \approx 131^\circ$$

Both angles are possible, since either one leaves room for the other two angles of the triangle. Use the cosine rule for $QR$ in each case:

$$QR^2 = 10^2 + 12^2 - 2(10)(12)\cos P$$

• If $P \approx 48.6^\circ$: $QR = 9.2333\ldots \approx 9.23$ cm.
• If $P \approx 131^\circ$: $QR = 20.068\ldots \approx 20.1$ cm.

Two quite different triangles, with exactly the same area.

Example 4: A field with four sides

A farmer’s field $ABCD$ has $AB = 50$ m, $BC = 70$ m and $\angle ABC = 110^\circ$. The fence $AD$ is $80$ m long and $\angle CAD = 35^\circ$. Find the area of the field.

Solution. Split the field along the diagonal $AC$ into two triangles.

Triangle $ABC$ (two sides and the included angle):

$$\text{Area}_1 = \frac{1}{2}(50)(70)\sin 110^\circ = 1644.46\ldots \text{ m}^2$$

For triangle $ACD$ we need $AC$. Use the cosine rule in triangle $ABC$:

$$AC = \sqrt{50^2 + 70^2 - 2(50)(70)\cos 110^\circ} = 98.965\ldots \text{ m}$$

Triangle $ACD$ has sides $AC$ and $AD = 80$ with the included angle $35^\circ$:

$$\text{Area}_2 = \frac{1}{2}(98.965\ldots)(80)\sin 35^\circ = 2270.56\ldots \text{ m}^2$$ $$\text{Total area} = 1644.46\ldots + 2270.56\ldots = 3915.03\ldots \approx 3920 \text{ m}^2 \quad (\text{3 s.f.})$$

Common mistakes

Using an angle that isn’t between the two sides. $\frac{1}{2}ab\sin C$ needs the angle enclosed by $a$ and $b$. In Example 2 the angle $B$ is opposite the $7$ cm side, so it’s enclosed by the $5$ and $8$ cm sides, and those are the ones you multiply.

Forgetting the second angle. When you find an angle from the area, $\sin^{-1}$ on your calculator only gives the acute angle. The obtuse angle $180^\circ - C$ gives the same sine and the same area. Check whether the question rules one out.

Calculator in radians. AA papers (and AI HL papers) assume radians unless a question says otherwise, so many students leave their GDC in radian mode. When a question gives an angle like $40^\circ$, switch to degree mode. In Example 1, radian mode gives $\frac{1}{2}(7)(9)\sin 40 \approx 23.5$, a wrong answer that doesn’t look wrong.

Rounding the side before using it again. In Example 4, using $AC \approx 99.0$ gives a total that can differ in the last figure. Store the full value of $AC$ in your calculator.

Dropping the one half. The formula is half of $ab\sin C$. Without the $\frac{1}{2}$ you get the area of a parallelogram with those two sides, which is twice too big.

Practice

1. (Warm-up) In triangle $ABC$, $a = 6$ cm, $b = 11$ cm and $C = 30^\circ$. Find the area.

Solution

$$\text{Area} = \frac{1}{2}(6)(11)\sin 30^\circ = 33 \times \frac{1}{2} = 16.5 \text{ cm}^2$$

2. (Warm-up) Find the exact area of an equilateral triangle with side $8$ cm.

Solution

All angles are $60^\circ$:

$$\text{Area} = \frac{1}{2}(8)(8)\sin 60^\circ = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3} \approx 27.7 \text{ cm}^2$$

3. (Core) In triangle $PQR$, $PQ = 14$ cm, $PR = 9$ cm and $\angle P = 125^\circ$. Find the area.

Solution

$$\text{Area} = \frac{1}{2}(14)(9)\sin 125^\circ = 51.606\ldots \approx 51.6 \text{ cm}^2 \quad (\text{3 s.f.})$$

The formula works for obtuse angles too, since $\sin 125^\circ = \sin 55^\circ \gt 0$.

4. (Core) A triangle has two sides of length $8$ cm and $9$ cm, and area $24 \text{ cm}^2$. The angle between these sides is obtuse. Find it.

Solution

$$\frac{1}{2}(8)(9)\sin\theta = 24 \quad\Rightarrow\quad \sin\theta = \frac{24}{36} = \frac{2}{3}$$

The acute solution is $\theta = 41.810\ldots^\circ$. The angle is obtuse, so

$$\theta = 180^\circ - 41.810\ldots^\circ = 138.189\ldots^\circ \approx 138^\circ \quad (\text{3 s.f.})$$

5. (Core) A triangle has sides $9$ cm, $10$ cm and $13$ cm. Find its area.

Solution

Find the angle $C$ opposite the $13$ cm side, between the $9$ and $10$ cm sides:

$$\cos C = \frac{9^2 + 10^2 - 13^2}{2(9)(10)} = \frac{12}{180} \quad\Rightarrow\quad C = 86.177\ldots^\circ$$$$\text{Area} = \frac{1}{2}(9)(10)\sin C = 44.899\ldots \approx 44.9 \text{ cm}^2 \quad (\text{3 s.f.})$$

6. (Core) In triangle $ABC$, $\angle A = 50^\circ$, $\angle B = 70^\circ$ and $c = AB = 12$ cm. Find the area of the triangle.

Solution

$C = 180^\circ - 50^\circ - 70^\circ = 60^\circ$. Use the sine rule to find $b = AC$:

$$\frac{b}{\sin 70^\circ} = \frac{12}{\sin 60^\circ} \quad\Rightarrow\quad b = \frac{12\sin 70^\circ}{\sin 60^\circ} = 13.020\ldots \text{ cm}$$

Now $b$ and $c$ enclose angle $A$:

$$\text{Area} = \frac{1}{2}bc\sin A = \frac{1}{2}(13.020\ldots)(12)\sin 50^\circ = 59.846\ldots \approx 59.8 \text{ cm}^2 \quad (\text{3 s.f.})$$

7. (Core) A triangular plot of land has two sides of $120$ m and $85$ m, with an angle of $72^\circ$ between them. Find its area in square metres and in hectares ($1$ hectare $= 10\,000 \text{ m}^2$).

Solution

$$\text{Area} = \frac{1}{2}(120)(85)\sin 72^\circ = 4850.38\ldots \approx 4850 \text{ m}^2 \quad (\text{3 s.f.})$$

In hectares: $4850.38\ldots \div 10\,000 \approx 0.485$ ha.

8. (Challenge) A triangle has sides $x$ cm and $(x + 2)$ cm with an angle of $150^\circ$ between them. Its area is $12 \text{ cm}^2$. Find $x$.

Solution

$\sin 150^\circ = \dfrac{1}{2}$, so

$$\frac{1}{2}x(x + 2)\left(\frac{1}{2}\right) = 12 \quad\Rightarrow\quad x(x + 2) = 48 \quad\Rightarrow\quad x^2 + 2x - 48 = 0$$$$(x + 8)(x - 6) = 0 \quad\Rightarrow\quad x = 6 \quad (\text{since } x \gt 0)$$

Check: $\frac{1}{2}(6)(8)\sin 150^\circ = 24 \times \frac{1}{2} = 12$. ✓

9. (Challenge) In triangle $ABC$, $AB = 10$ cm, $AC = 7$ cm, and the area is $28 \text{ cm}^2$. Find the two possible exact lengths of $BC$.

Solution

$$\frac{1}{2}(10)(7)\sin A = 28 \quad\Rightarrow\quad \sin A = \frac{28}{35} = \frac{4}{5}$$

Then $\cos^2 A = 1 - \dfrac{16}{25} = \dfrac{9}{25}$, so $\cos A = \dfrac{3}{5}$ (acute $A$) or $\cos A = -\dfrac{3}{5}$ (obtuse $A$). By the cosine rule:

$$BC^2 = 10^2 + 7^2 - 2(10)(7)\cos A = 149 - 140\cos A$$

• Acute: $BC^2 = 149 - 84 = 65$, so $BC = \sqrt{65} \approx 8.06$ cm.
• Obtuse: $BC^2 = 149 + 84 = 233$, so $BC = \sqrt{233} \approx 15.3$ cm.