Area with Sine

The height is the missing piece

The area of every triangle is ½ × base × height, where the height is the perpendicular distance from the base to the opposite corner. That holds whether or not the triangle has a right angle.

The difficulty comes when the information given is two sides and the angle between them: sides a and b, with the angle C where they meet. Side a can be the base, but the height onto it is not b. Side b runs at a slant, and the height is the perpendicular, which is shorter.

The base stays fixed and the top corner slides along a rail parallel to it. The slanting sides change length, but the height between the rails does not, so the area ½ × 6 × 4 = 12 does not change. The area depends on the height, not on the length of a slanting side.

Drop the height inside the triangle

Draw the height h from the corner A, perpendicular to side a. It meets side a at a right angle and splits the triangle into two right triangles.

Look at the right triangle that contains the angle C. Its hypotenuse is the side b, because b is opposite its right angle. The height h is the side opposite the angle C.

The height h dropped from A onto side a. In the right triangle on the left, b is the hypotenuse and h is opposite the angle C.

The height in terms of b and C

In that right triangle, the sine of C is the opposite side over the hypotenuse: $\sin C=h/b$sin C = h / b.

Multiply both sides by b: h = b sin C. The height is found from the two things that are known, the side b and the angle C, without measuring anything.

Area = ½ab sin C

Put that height into the area formula, with a as the base: Area = ½ × a × h = ½ × a × b sin C. Written more compactly, Area = ½ab sin C.

In words: the area is half the product of two sides and the sine of the angle between them. The angle must be the one between the two sides, the angle where they meet.

The letters only name the parts. The same argument works with any side as the base, so Area = ½ab sin C = ½bc sin A = ½ca sin B. Each version uses two sides and the angle between those two.

A worked value

Sides of 7 cm and 5 cm meet at an angle of 52°. The area is ½ × 7 × 5 × sin 52°. With sin 52° = 0.7880, to 4 decimal places, that is 17.5 × 0.7880 = 13.79, so the area is 13.8 cm², to 1 decimal place.

Check through the height: h = 5 sin 52° = 5 × 0.7880 = 3.940 cm, and ½ × 7 × 3.940 = 13.79 cm². The two routes are the same calculation, done in a different order.

Sides of 7 cm and 5 cm with 52° between them. The height is 5 sin 52° = 3.94 cm, and the area is ½ × 7 × 5 × sin 52° = 13.8 cm², to 1 decimal place.

Two special cases

When the angle between the two sides is 90°, sin 90° = 1 and the formula becomes ½ab × 1 = ½ab. That is the familiar area of a right triangle: the two sides that meet at the right angle are the base and the height.

An equilateral triangle with sides of 6 has every angle 60°, and $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$sin 60° = √3/2. Its area is $\frac{1}{2}\times 6\times 6\times \frac{\sqrt{3}}{2}=9\sqrt{3}$½ × 6 × 6 × √3/2 = 9√3, which is 15.59 to 2 decimal places.

When the angle is obtuse

If the angle C is obtuse, the height from A falls outside the triangle, onto side a extended past C. The right triangle it makes contains the angle 180° − C, not C, so the height is b sin (180° − C).

An angle and its supplement have the same sine: sin (180° − C) = sin C. So the height is still b sin C, and Area = ½ab sin C holds for an obtuse angle too. A calculator gives the sine of an obtuse angle as a positive number, so the area comes out positive, as an area must.

For example, sides of 80 m and 110 m with 150° between them give an area of ½ × 80 × 110 × sin 150°. Since sin 150° = sin 30° = ½, the area is $\frac{1}{2}\times 80\times 110\times \frac{1}{2}=2200m^2$½ × 80 × 110 × ½ = 2200 m².

Sides of 80 m and 110 m meeting at 150°. The triangle is long and thin, and its area is $\frac{1}{2}\times 80\times 110\times \sin 150^{\circ }=2200m^2$½ × 80 × 110 × sin 150° = 2200 m².

Running the formula backwards

The formula links four quantities: two sides, the angle between them and the area. Given any three, it gives the fourth.

A triangle has an area of 24 cm², one side of 8 cm, and an angle of 30° between that side and an unknown side a. Then 24 = ½ × a × 8 × sin 30° = ½ × a × 8 × ½ = 2a, so a = 12 cm.

A triangle has sides of 6 cm and 10 cm and an area of 15 cm². Then 15 = ½ × 6 × 10 × sin C = 30 sin C, so sin C = ½. Two angles between 0° and 180° have a sine of ½: 30° and 180° − 30° = 150°. Both make a triangle with these two sides and this area, one with an acute angle between the sides and one with an obtuse angle.

The usual mistakes

Using an angle that is not between the two sides. In ½ab sin C, the angle C is where a and b meet. If the given angle is somewhere else, find the angle between the two sides first.

Forgetting the ½. The product ab sin C is the area of a parallelogram with sides a and b; the triangle is half of it.

Leaving out the sine. Half the product of the two sides, ½ab, is the area only when the angle between them is 90°.

Taking the sine of an obtuse angle as negative. The sine of 150° is +½, the same as the sine of 30°.