Area of Oblique Triangles

The area of an oblique triangle (a triangle with no right angle) can be found without knowing its height. If you know two sides and the angle between them, use \text{Area} = \frac{1}{2}bc \sin A. If you know all three sides, use Heron’s formula: \text{Area} = \sqrt{s(s - a)(s - b)(s - c)}, where s is half the perimeter.

Before this lesson: You should know the Area of Triangles, the Law of Sines and the Law of Cosines.

Why the Sine Gives the Area

The usual formula is \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}. In an oblique triangle, you often don’t know the height. But you can build it from a side and an angle.

In the picture, the height h makes a right triangle with side b. In that right triangle, \sin A = \frac{h}{b}, so h = b \sin A. Put that height into the usual formula with base c, and you get \frac{1}{2} c \cdot b \sin A.

Area Formulas for Oblique Triangles

Two sides and the angle between them (SAS)

\text{Area} = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}{2}ab \sin C

Pick the version whose angle sits between the two sides you know.

Three sides (SSS): Heron’s formula

s = \frac{a + b + c}{2}, \quad \text{Area} = \sqrt{s(s - a)(s - b)(s - c)}

Here s is the semiperimeter, half the distance around the triangle. Heron’s formula needs no angles at all.

Other cases

For ASA, AAS or SSA, first use the law of sines to find a second side. Then you have two sides and the angle between them.

Worked Examples

Use a calculator in degree mode. Round areas to the nearest hundredth.

Example 1: Two sides and the angle between them

In triangle ABC, b = 9, c = 12 and A = 40^\circ. Find the area.

Angle A is between sides b and c, so use \text{Area} = \frac{1}{2}bc \sin A

\text{Area} = \frac{1}{2}(9)(12) \sin 40^\circ

\text{Area} = 54 \sin 40^\circ

\text{Area} \approx 54(0.6428)

\text{Area} \approx 34.71

Answer: about 34.71 square units

Example 2: An exact area with an obtuse angle

In triangle ABC, a = 6, b = 10 and C = 150^\circ. Find the exact area.

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

\text{Area} = \frac{1}{2}(6)(10) \sin 150^\circ

\text{Area} = 30 \sin 150^\circ

\sin 150^\circ = \sin 30^\circ = \frac{1}{2}

\text{Area} = 30 \cdot \frac{1}{2} = 15

Answer: exactly 15 square units. The sine of an obtuse angle is positive, so the formula still works.

Example 3: Heron’s formula

Find the area of a triangle with sides 7, 8 and 9.

s = \frac{7 + 8 + 9}{2} = \frac{24}{2} = 12

Find each difference: s - a = 12 - 7 = 5, s - b = 12 - 8 = 4, s - c = 12 - 9 = 3

\text{Area} = \sqrt{12 \cdot 5 \cdot 4 \cdot 3}

\text{Area} = \sqrt{720}

\sqrt{720} = \sqrt{144 \cdot 5} = 12\sqrt{5}

12\sqrt{5} \approx 26.83

Answer: 12\sqrt{5} \approx 26.83 square units

Example 4: Two angles and a side (ASA)

In triangle ABC, A = 35^\circ, B = 75^\circ and c = 20. Find the area.

C = 180^\circ - 35^\circ - 75^\circ = 70^\circ

b = \frac{20 \sin 75^\circ}{\sin 70^\circ} \approx \frac{20(0.9659)}{0.9397} \approx 20.558

\text{Area} = \frac{1}{2}bc \sin A

\text{Area} \approx \frac{1}{2}(20.558)(20) \sin 35^\circ

\text{Area} \approx 205.58 \sin 35^\circ

\text{Area} \approx 205.58(0.5736) \approx 117.92

Answer: about 117.92 square units

Example 5: Word problem with a triangular lot

A triangular lot has sides of 120 ft, 150 ft and 200 ft. Sod costs $0.45 per square foot. About how much will it cost to cover the lot?

s = \frac{120 + 150 + 200}{2} = \frac{470}{2} = 235

Find each difference: 235 - 120 = 115, 235 - 150 = 85, 235 - 200 = 35

\text{Area} = \sqrt{235 \cdot 115 \cdot 85 \cdot 35}

\text{Area} = \sqrt{80{,}399{,}375}

\text{Area} \approx 8966.57

8966.57 \cdot 0.45 \approx 4034.96

Answer: about $4,034.96

Common Mistakes

Related Methods

Parallelograms

A parallelogram is two copies of the same triangle. With sides a and b and an angle \theta between them, its area is ab \sin \theta, with no one half.

Finding an angle from the area

If you know the area and two sides, solve the formula for the sine. For example, 30 = \frac{1}{2}(8)(10) \sin C gives \sin C = 0.75.

Right triangles

If the angle is 90^\circ, \sin 90^\circ = 1, and the formula becomes \frac{1}{2}ab, the usual area of a right triangle.

Practice Problems

Round answers to the nearest hundredth unless the problem asks for an exact value.

1. Find the area of triangle ABC with a = 8, b = 11 and C = 30^\circ.
2. Find the area of triangle ABC with a = 14, c = 9 and B = 72^\circ.
3. Use Heron’s formula to find the exact area of a triangle with sides 5, 6 and 7.
4. Use Heron’s formula to find the area of a triangle with sides 13, 14 and 15.
5. Find the exact area of triangle ABC with b = 5, c = 8 and A = 120^\circ.
6. A parallelogram has sides 7 and 12 with a 65^\circ angle between them. Find its area.
7. A triangle has area 30, a = 8 and b = 10, and angle C is acute. Find C to the nearest tenth of a degree.
8. Find the area of an isosceles triangle with sides 10, 10 and 12.
9. In triangle ABC, A = 50^\circ, B = 60^\circ and a = 10. Find the area.
10. A sail is a triangle with sides of 12 ft and 15 ft that meet at a 70^\circ angle. Find the area of the sail.

\text{Area} = \frac{1}{2}(8)(11) \sin 30^\circ

\text{Area} = 44 \sin 30^\circ

\text{Area} = 22

\text{Area} = \frac{1}{2}(14)(9) \sin 72^\circ

\text{Area} = 63 \sin 72^\circ

\text{Area} \approx 63(0.9511) \approx 59.92

s = \frac{5 + 6 + 7}{2} = 9

\text{Area} = \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = \sqrt{216}

\sqrt{216} = \sqrt{36 \cdot 6} = 6\sqrt{6}

s = \frac{13 + 14 + 15}{2} = 21

\text{Area} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056}

\sqrt{7056} = 84

\text{Area} = \frac{1}{2}(5)(8) \sin 120^\circ

\text{Area} = 20 \sin 120^\circ

\text{Area} = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3}

\text{Area} = (7)(12) \sin 65^\circ

\text{Area} = 84 \sin 65^\circ

\text{Area} \approx 84(0.9063) \approx 76.13

30 = \frac{1}{2}(8)(10) \sin C

30 = 40 \sin C

\sin C = 0.75

C = \sin^{-1}(0.75) \approx 48.6^\circ

s = \frac{10 + 10 + 12}{2} = 16

\text{Area} = \sqrt{16 \cdot 6 \cdot 6 \cdot 4} = \sqrt{2304}

\sqrt{2304} = 48

C = 180^\circ - 50^\circ - 60^\circ = 70^\circ

b = \frac{10 \sin 60^\circ}{\sin 50^\circ} \approx \frac{10(0.8660)}{0.7660} \approx 11.305

\text{Area} \approx \frac{1}{2}(10)(11.305) \sin 70^\circ

\text{Area} \approx 56.525(0.9397) \approx 53.12

\text{Area} = \frac{1}{2}(12)(15) \sin 70^\circ

\text{Area} = 90 \sin 70^\circ

\text{Area} \approx 90(0.9397) \approx 84.57

FAQ

How do you find the area of a triangle without the height?

Use two sides and the angle between them: \text{Area} = \frac{1}{2}ab \sin C. If you know all three sides, use Heron’s formula.

What is Heron’s formula?

It finds a triangle’s area from its three sides. Let s be half the perimeter; then the area is \sqrt{s(s - a)(s - b)(s - c)}.

Does the sine area formula work for obtuse angles?

Yes. The sine of an angle between 90^\circ and 180^\circ is positive, so the area comes out positive.

What is an oblique triangle?

It is any triangle without a right angle. It can be acute, with all angles under 90^\circ, or obtuse, with one angle over 90^\circ.

Summary

• SAS: \text{Area} = \frac{1}{2}ab \sin C, using the angle between the two sides.
• SSS: Heron’s formula, \sqrt{s(s - a)(s - b)(s - c)} with s = \frac{a + b + c}{2}.
• ASA, AAS or SSA: find a second side with the law of sines first.
• A parallelogram’s area is ab \sin \theta.
• Keep extra digits until the last step.

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