Area of a triangle using sine — half ab sinC — Mathematics

Before this

Area of triangles, parallelograms and trapeziums · coming soon
The sine rule — finding sides and angles
The cosine rule — SAS and SSS triangles

This leads to

3D trigonometry
Arc length and sector area
What is a vector — magnitude and direction

A field with no height

You have known the area of a triangle for years: half the base times the height. It is a good formula. But it has a weakness that nobody mentions. It needs the height, and the height is usually the one length nobody gives you.

Here is a real kind of problem. This example is invented. A farmer owns a triangular field. From the corner at the gate, A, one fence runs 80 m to B and another runs 65 m to C. The angle between the two fences at the gate is 50°. The third side of the field runs along a river.

How big is the field? The height would be the distance from C straight across to the fence AB, at exactly 90°. That line runs through the middle of the crops, and nobody can measure it there.

What he does have is two sides and the angle between them. The cosine rule lecture showed that these three facts fix the whole triangle, so they must also fix its area. This lecture finds it, with one idea you already own.

The height is hiding in a right-angled triangle

Draw the triangle and drop the height $h$ from C to the point H on AB, as you did to prove the sine rule. The height cuts off a right-angled triangle on the left, ACH.

The height h from C meets AB at a right angle at H. In the right-angled triangle ACH, the side b is the hypotenuse and h is opposite angle A, so h = b sin A.

Look at the triangle ACH on its own. Its hypotenuse is $b$ , the side AC. The height $h$ is opposite angle A. So, from the definition of sine:

sin A = \frac{h}{b}, so h = b sin A

This is the whole trick. We did not measure the height. We calculated it from a side we know and an angle we know. Now put this height into the formula you already trust:

Area = \frac{1}{2} \times c \times h

Replace $h$ with $b sin A$ :

Area = \frac{1}{2} \times c \times b sin A

Written in the usual order, the area is $\frac{1}{2} b c sin A$ . The sides $b$ and $c$ are the two sides that meet at A. Angle A is the angle between them.

That is it. That is the whole idea. Half the base times the height, with the height written as a side times a sine. The new formula is the old formula in disguise.

Back to the field

For the farmer, $b = 65$ m, $c = 80$ m and A = 50°, the angle between them.

Area = \frac{1}{2} \times 80 \times 65 \times sin 5 0^{\circ}

Here $\frac{1}{2} \times 80 \times 65 = 2600$ , and $sin 5 0^{\circ} \approx 0.7660$ . So the area is about $2600 \times 0.7660 \approx 1990$ m², to 3 significant figures (3 s.f.).

Check. At 90°, the field would have area 2600 m². At 50°, the fence AC leans towards AB, so the field is thinner. A smaller area makes sense. ✓

Stop and think

A student uses the same field but writes the area as $\frac{1}{2} \times 80 \times 65 \times sin B$ , using the angle at B instead. Why does this not work?

Show the answer

The height came from the right-angled triangle at A. In that triangle, the side $b$ and the angle A belong together: $h = b sin A$ . Angle B is not between the sides 80 m and 65 m. It sits at the end of the 80 m side, between AB and BC. So $65 sin B$ is not the height of anything. The rule needs the angle where the two sides you are using meet.

One rule, from any corner

Nothing in the argument depended on which corner we called A. The height could have come down from any corner. So the rule works with any two sides and the angle between them. Books call this angle the included angle: it is included between the two sides.

Area of a triangle

The area of any triangle is half the product of two sides times the sine of the angle between them.

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

or, from the other two corners, $\frac{1}{2} b c sin A$ or $\frac{1}{2} c a sin B$ .

Side $a$ faces A, side $b$ faces B, side $c$ faces C. In $\frac{1}{2} ab sin C$ , sides $a$ and $b$ both touch corner C, the one corner neither of them faces. In words: two sides, and the angle where they meet.

This is the case books call SAS, as in the cosine rule lecture. The cosine rule found the third side from SAS. This formula finds the area from exactly the same three facts.

What happens when the angle is obtuse?

The proof assumed that the height lands between A and B. That is true when A is acute. Before reading the case where it is not, test your sense of what should happen.

Predict — which triangle is bigger?

Two triangles both have sides of 6 cm and 8 cm meeting at A. In the first, the angle between them is 30°. In the second, it is 150°. Which has the larger area?

This box asks which of two triangles is bigger. Both have sides of 6 cm and 8 cm meeting at A. The angle between them is 30° in one and 150° in the other.

They have exactly the same area, 12 cm². In both, the 6 cm side rises to a height of 3 cm above the 8 cm base. In the second triangle it leans backwards, but it reaches the same height. The figure in the next section shows both triangles.

The surprise is real: a 30° angle and a 150° angle give the same area. Here are the two triangles. Look at the heights.

Both triangles have base 8 cm and a 6 cm side. With 30° at A, C₁ is 3 cm above the base. With 150° at A, the side leans back over the extended base line, and C₂ is also 3 cm above it. Same base, same height, same area: 12 cm².

Now the general case. This is the hard part of the lecture. Spend time here. When A is obtuse, the height from C lands outside the triangle, on the base line extended beyond A.

When angle A is obtuse, the height h from C lands at H on the base line extended beyond A. In the right-angled triangle ACH, the angle at A is 180° − A, so h = b sin(180° − A).

The right-angled triangle ACH still exists. But its angle at A is not A. It is $18 0^{\circ} - A$ , because the two angles at A sit on a straight line and add up to 180°. So the height is

h = b sin (18 0^{\circ} - A)

The area is still half the base times the height, so it is $\frac{1}{2} b c sin (18 0^{\circ} - A)$ . The formula $\frac{1}{2} b c sin A$ gives the same answer only if $sin A = sin (18 0^{\circ} - A)$ .

The sine rule lecture met exactly this fact. For any angle $θ$ between 0° and 180°, $sin θ = sin (18 0^{\circ} - θ)$ . That is how the sine of an obtuse angle is defined, and it is what your calculator does: $sin 15 0^{\circ} = 0.5 = sin 3 0^{\circ}$ . So no adjustment is needed. The formula works for every triangle, acute, right-angled or obtuse.

The right angle is a check, not a separate case. If A = 90°, then $sin A = 1$ , and the formula becomes $\frac{1}{2} b c$ . The two sides are now the base and the height, and you are back to the formula from your first year of secondary school. ✓

Which angle gives the biggest triangle?

Keep the two sides at 6 cm and 8 cm, joined at A like a hinge. Open the hinge from almost closed to almost flat. The area is $\frac{1}{2} \times 6 \times 8 \times sin A = 24 sin A$ . Watch how it changes.

Explore — open the hinge and watch the area

Side AC = 6 cm swings about A. Side AB = 8 cm stays on the base. Change the angle A. Watch the height of C and the area of the triangle.

Angle A: 40°

This table shows the area of a triangle with sides 6 cm and 8 cm, for different angles A between them. The height of C above the base is 6 sin A, and the area is 24 sin A.

Angle A	$sin A$	Height (cm)	Area (cm²)
10°	≈ 0.1736	≈ 1.04	≈ 4.17
30°	0.5	3	12
60°	≈ 0.8660	≈ 5.20	≈ 20.78
90°	1	6	24
120°	≈ 0.8660	≈ 5.20	≈ 20.78
150°	0.5	3	12
170°	≈ 0.1736	≈ 1.04	≈ 4.17

The area rises to its largest value, 24 cm², at 90°, then falls again. Angles that add up to 180° give the same area.

You should have seen three things. The area grows as the hinge opens, reaches 24 cm² at exactly 90°, and then shrinks again. Second, every angle has a twin: 60° and 120° give the same area, and so do 30° and 150°. Third, a nearly closed or nearly flat hinge gives almost no area at all. All three are the sine curve between 0° and 180°: it rises to 1 at 90° and falls back symmetrically.

So two fixed lengths of fencing from a corner post enclose the most ground when they meet at a right angle.

One area, three ways: the sine rule appears again

Here is a connection that is worth slowing down for. The theorem box wrote the same area in three ways. They are three formulas for one number, so they are equal:

\frac{1}{2} b c sin A = \frac{1}{2} c a sin B = \frac{1}{2} ab sin C

Now divide every part by $\frac{1}{2} ab c$ . In the first part, the $b$ and the $c$ cancel, and $\frac{sin A}{a}$ is left. The same happens in the other two parts.

\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}

That is the sine rule, in the form used to find an angle. The sine rule lecture proved it with one height written two ways. Here it falls out of the area in one line. This is no accident. Both proofs use the same height, $h = b sin A$ .

Worked examples

Worked Example 1 — Two sides and the angle between them

In triangle ABC, $a = 8$ cm, $b = 11$ cm and C = 42°. Find the area.

Think first. Sides $a$ and $b$ both touch corner C, so C is the angle between them. The formula from corner C fits directly.

Area = \frac{1}{2} \times 8 \times 11 \times sin 4 2^{\circ}

Here $\frac{1}{2} \times 8 \times 11 = 44$ . So the area is about $44 \times 0.6691 \approx 29.4$ cm².

Check. The largest possible area with these sides is 44 cm², at 90°. Here $sin 4 2^{\circ}$ is about two-thirds, so about two-thirds of 44 is right. ✓

Worked Example 2 — An obtuse angle

In triangle ABC, AB = 14 cm, AC = 10 cm and A = 130°. Find the area.

Think first. AB and AC meet at A, so 130° is the angle between them. It is obtuse, but the formula needs no change.

Area = \frac{1}{2} \times 10 \times 14 \times sin 13 0^{\circ}

Here $\frac{1}{2} \times 10 \times 14 = 70$ . The calculator gives $sin 13 0^{\circ} \approx 0.7660$ , which is $sin 5 0^{\circ}$ . So the area is about $70 \times 0.7660 \approx 53.6$ cm².

Check. The twin angle is $18 0^{\circ} - 13 0^{\circ} = 5 0^{\circ}$ . A triangle with the same two sides and 50° between them must have the same area, and $70 sin 5 0^{\circ}$ gives the same 53.6 cm². ✓

Worked Example 3 — Working backwards to a side

Triangle PQR has area 40 cm². The angle at R is 65°, and RP = 9 cm. Find RQ.

Think first. The sides RP and RQ both meet at R, so 65° is the angle between them. Call the unknown side RQ = $x$ . Write the area formula with $x$ in it, then solve.

40 = \frac{1}{2} \times 9 \times x \times sin 6 5^{\circ}

Here $\frac{1}{2} \times 9 \times sin 6 5^{\circ} \approx 4.0784$ . So $40 \approx 4.0784 x$ , and $x \approx 40 \div 4.0784 \approx 9.81$ cm.

Check. Put it back: $\frac{1}{2} \times 9 \times 9.808 \times sin 6 5^{\circ} \approx 40.0$ cm². ✓

Worked Example 4 — Working backwards to an angle

A triangle has sides 8 cm and 10 cm and an area of 30 cm². Find the angle between the two sides.

Think first. Call the angle $θ$ . This time the sine is the unknown.

30 = \frac{1}{2} \times 8 \times 10 \times sin θ

Here $\frac{1}{2} \times 8 \times 10 = 40$ , so $30 = 40 sin θ$ . So $sin θ = 30 \div 40 = 0.75$ . The calculator gives $θ = sin^{- 1} (0.75) \approx 48. 6^{\circ}$ .

Now remember the twins. The angle $18 0^{\circ} - 48. 6^{\circ} = 131. 4^{\circ}$ has the same sine, so it gives the same area. Both triangles exist. With 48.6°, the third side is about 7.63 cm. With 131.4°, the third side is about 16.4 cm (both from the cosine rule). The question has two answers: $θ \approx 48. 6^{\circ}$ or $θ \approx 131. 4^{\circ}$ .

Check. $40 sin 48. 6^{\circ} \approx 30.0$ and $40 sin 131. 4^{\circ} \approx 30.0$ . ✓ This is the same test as the ambiguous case of the sine rule: whenever you find an angle from its sine, ask whether 180° minus that angle also fits.

Worked Example 5 — Only three sides

A triangle has sides 3 cm, 5 cm and 7 cm. Find its area exactly.

Think first. There is no angle yet. The cosine rule finds one, and then the area formula can use it. Find the angle between the sides 3 cm and 5 cm, which is the angle facing the 7 cm side.

Here $3^{2} + 5^{2} - 7^{2} = 9 + 25 - 49 = - 15$ , and $2 \times 3 \times 5 = 30$ .

cos θ = \frac{- 15}{30} = - 0.5

So $θ = 12 0^{\circ}$ , an obtuse angle. Its twin is 60°, so $sin 12 0^{\circ} = sin 6 0^{\circ} = \frac{3}{2}$ , from the exact values lecture.

Area = \frac{1}{2} \times 3 \times 5 \times \frac{3}{2}

Area = \frac{15 3}{4} \approx 6.50 cm^{2}

Check. The calculator gives $7.5 sin 12 0^{\circ} \approx 6.495$ . ✓ With the cosine rule, the sine rule and this formula, two sides and the angle between them give you everything else.

Three traps

Trap 1 — An angle that is not between the two sides

In Worked Example 1, angle A is about 46.6°. Using it with the sides 8 and 11 gives $\frac{1}{2} \times 8 \times 11 \times sin 46. 6^{\circ} \approx 32.0$ cm². That is wrong: the true area is 29.4 cm². Angle A sits at the end of side $b$ , not where $a$ and $b$ meet. Before you write the formula, put your finger on the corner where your two sides touch. That corner's angle is the one you use.

Trap 2 — Forgetting the half

The formula is half the base times the height, so it keeps the half: $\frac{1}{2} ab sin C$ . Without the half, $ab sin C$ is the area of a parallelogram with sides $a$ and $b$ , which is two copies of the triangle. If an answer is exactly double what you expected, look for the missing half.

Trap 3 — One answer when there are two

Finding an angle from an area means finding an angle from its sine. The calculator gives only the acute angle. Always test the twin, 180° minus that angle. If the question gives no other information, both are possible, as in Worked Example 4. If it says the angle is obtuse, or gives the third side, only one fits.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A triangle has sides 6 cm and 10 cm, with an angle of 30° between them. Find its area exactly. Then find the area if the angle is 150° instead.

Hint 1
Use half the product of the two sides times the sine of the angle between them.

Hint 2
$sin 3 0^{\circ} = \frac{1}{2}$ exactly. What is $sin 15 0^{\circ}$ ?

Full answer
Area $= \frac{1}{2} \times 6 \times 10 \times sin 3 0^{\circ} = 30 \times \frac{1}{2} = 15$ cm², exactly. With 150°, $sin 15 0^{\circ} = sin (18 0^{\circ} - 15 0^{\circ}) = sin 3 0^{\circ} = \frac{1}{2}$ , so the area is also 15 cm². Check: the height above the 10 cm side is $6 sin 3 0^{\circ} = 3$ cm in both cases, and $\frac{1}{2} \times 10 \times 3 = 15$ . ✓
A parallelogram has sides 9 cm and 14 cm, and one of its angles is 70°. Find its area.

Hint 1
A diagonal cuts a parallelogram into two triangles. Are they the same size?

Hint 2
Each triangle has sides 9 cm and 14 cm with 70° between them.

Full answer
A diagonal cuts the parallelogram into two congruent triangles, each with sides 9 cm and 14 cm and the 70° angle between them. One triangle has area $\frac{1}{2} \times 9 \times 14 \times sin 7 0^{\circ}$ . Two of them give $9 \times 14 \times sin 7 0^{\circ} = 126 sin 7 0^{\circ} \approx 118$ cm² (3 s.f.). So a parallelogram's area is $ab sin θ$ : the formula without the half. Check: the other angle of the parallelogram is 110°, and $sin 11 0^{\circ} = sin 7 0^{\circ}$ , so the other diagonal gives the same answer. ✓
A triangle has sides 5 cm, 7 cm and 8 cm. Find the angle facing the 7 cm side, then the exact area.

Hint 1
Use the cosine rule in its angle form. The side facing the angle you want takes the minus sign.

Hint 2
$cos θ = \frac{5 ^{2} + 8 ^{2} - 7 ^{2}}{2 \times 5 \times 8}$ . The answer is an exact value.

Full answer
$cos θ = \frac{25 + 64 - 49}{80} = \frac{40}{80} = 0.5$ , so $θ = 6 0^{\circ}$ . The angle is between the sides 5 cm and 8 cm, so the area is $\frac{1}{2} \times 5 \times 8 \times sin 6 0^{\circ} = 20 \times \frac{3}{2} = 103 \approx 17.3$ cm². Check with Heron's formula, which uses only the three sides. Half the perimeter is $s = 10$ , and $s (s - 5) (s - 7) (s - 8) = 10 \times 5 \times 3 \times 2 = 300 = 103$ . ✓
A stretch. A regular hexagon is drawn inside a circle of radius 10 cm, with its six corners on the circle. (a) Find the area of the hexagon. (b) Do the same for a regular 12-sided shape and a regular 360-sided shape in the same circle. (c) What number are these areas getting closer to, and why?

Hint 1
Join the centre to every corner. This cuts the shape into identical triangles, each with two sides of 10 cm.

Hint 2
With $n$ corners, the angle at the centre of each triangle is $36 0^{\circ} \div n$ . So each triangle has area $\frac{1}{2} \times 10 \times 10 \times sin (36 0^{\circ} \div n)$ .

Full answer
(a) Six triangles, each with two 10 cm sides and 60° between them. The area is $6 \times \frac{1}{2} \times 100 \times sin 6 0^{\circ} = 300 \times \frac{3}{2} = 1503 \approx 260$ cm². (b) For 12 sides, each angle is 30°, so the area is $12 \times 50 \times sin 3 0^{\circ} = 300$ cm² exactly. For 360 sides, each angle is 1°, so the area is $360 \times 50 \times sin 1^{\circ} \approx 314.14$ cm². (c) The shapes fill the circle more and more closely, so their areas approach the area of the circle, $π \times 1 0^{2} = 100 π \approx 314.16$ cm². Check: each area is bigger than the last and smaller than $100 π$ , as it must be for shapes inside the circle. ✓ Archimedes used shapes like these, with up to 96 sides, to trap the value of $π$ .

The bigger idea

Mathematicians have wanted the area of a triangle from its measurable parts for two thousand years. Heron of Alexandria, who worked in the first century AD, proved a formula for the area from the three sides alone in his book Metrica. Historians think Archimedes may have known it earlier. You met Heron's formula as a check in Try it yourself. Today it is usually proved with the cosine rule and the formula of this lecture together, because both describe the same fixed triangle.

The formula also carries a deeper idea. The quantity $ab sin θ$ measures how much area two lengths can enclose when they point in different directions. It is zero when they point the same way, and largest when they are at right angles. In the vectors lectures, two arrows will play the part of the two sides. At Pre-University this quantity becomes a way of combining them, used in physics for the turning effect of a force.

Key concepts in this lesson

The area of any triangle is $\frac{1}{2} ab sin C$ : half the product of two sides times the sine of the angle between them.
The formula is half the base times the height, with the height written as a side times a sine, $h = b sin A$ .
The angle must be the included angle, where the two sides meet; any other angle gives a wrong answer.
The formula works for obtuse angles because $sin θ = sin (18 0^{\circ} - θ)$ , so an angle and its twin give the same area.
With two fixed sides, the area is largest when the angle between them is 90°.
Writing the same area from all three corners and dividing by $\frac{1}{2} ab c$ gives the sine rule.