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Precalculus · G.SRT.D.9

Area from two sides and the included angle

The area of a triangle is half its base times its height. What if you know two sides and the angle between them, but not the height?

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Law of sinesPrecalculus
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The cross productPrecalculus
Where you are 6 in Laws of sines and cosines
Trigonometry7
Laws of sines and cosines6
Law of sines
Area from two sides and the included angle
The ambiguous caseLaw of cosinesHeron's formulaApplications of oblique triangles

When you reach for it

The area of a triangle is \tfrac{1}{2}(\text{base})(\text{height}). The base is usually a side you know; the height usually is not. When you know two sides and the angle between them (SAS), the height is one sine away, and the area follows without solving the rest of the triangle.

Any triangle whose parts you can find leads here too. AAS and ASA triangles give their missing sides by the law of sines, and SSS triangles give an angle by the law of cosines. Once two sides and the angle between them are known, this move gives the area.

As in those lessons, the triangle is ABC, and each side is named by the lowercase letter of the angle across from it.

The procedure

The height is a side times a sine

C B A D a b h
Figure 1Side CB is the base. The height from A is the side opposite C in the right triangle CDA, so h = b\sin C.

Take side CB, of length a, as the base. Drop the height from A to the line CB, meeting it at D, and call its length h. If C is acute and D falls on the base, then CDA is a right triangle with hypotenuse CA = b and the side h opposite the angle C. The sine ratio gives \sin C = \dfrac{h}{b}, so h = b\sin C. Then

\begin{aligned} \text{Area} &= \frac{1}{2}(\text{base})(\text{height}) \\ &= \frac{1}{2}a\,(b\sin C) = \frac{1}{2}ab\sin C. \end{aligned}

This is the same altitude that proved the law of sines; there it was compared with a second expression for itself, and here it is multiplied by the base.

When the included angle is obtuse

180^\circ - C
 C B A D a b h
Figure 2When C is obtuse the foot of the height lands outside the triangle, and the right triangle CDA holds 180^\circ - C. Its sine is \sin C, so h = b\sin C still.

If C is obtuse, the height from A lands on the line BC extended past C, outside the triangle. The base is still CB = a, because the area formula measures the height to the line through the base. The right triangle CDA now holds the angle between CA and the extension, and the two angles at C make a straight line, so that angle is 180^\circ - C. Then

h = b\sin(180^\circ - C) = b\sin C,

by the identity \sin(180^\circ - x) = \sin x from Unit circle symmetry and even-odd identities: the terminal points of x and 180^\circ - x are mirror images across the y-axis and share their y-coordinate. The area is \tfrac{1}{2}ab\sin C again.

If C = 90^\circ, the height from A is side b itself, and the formula gives \tfrac{1}{2}ab\sin 90^\circ = \tfrac{1}{2}ab: half the product of the legs, the right triangle’s area. All three cases agree. In the coordinate picture of the law of cosines, with C at the origin and B on the positive x-axis, A = (b\cos C,\ b\sin C), and the height is simply the y-coordinate of A.

The area formula

Any side can serve as the base, so in any triangle ABC,

\begin{aligned} \text{Area} &= \frac{1}{2}bc\sin A \\ &= \frac{1}{2}ac\sin B \\ &= \frac{1}{2}ab\sin C. \end{aligned}

In words: half the product of two sides times the sine of the angle between them. The angle must be the included one, the angle at the vertex where the two sides meet.

Same two sides, supplementary angles

Because \sin C = \sin(180^\circ - C), two triangles with the same two sides and included angles C and 180^\circ - C have the same area. They are not congruent: their third sides differ, as the law of cosines shows, since \cos(180^\circ - C) = -\cos C. The height is the same, and so is the base.

Working backward from an area

The formula has four quantities, so any three give the fourth.

A side. Given the area K, one side a and the included angle C, the other side is b = \dfrac{2K}{a\sin C}. Division is allowed because \sin C > 0 for 0^\circ < C < 180^\circ, and there is exactly one answer.

An angle. Given K and both sides,

\sin C = \frac{2K}{ab}.

The right side is positive. If it is less than 1, two angles between 0^\circ and 180^\circ have that sine: the acute angle C_1 that the inverse sine returns, and 180^\circ - C_1. Both are possible, because two sides with any angle between them close into a triangle. If it equals 1, only C = 90^\circ works, and this is the largest area two sides of lengths a and b can enclose, \tfrac{1}{2}ab. If it is more than 1, no triangle has that area.

Keep full calculator values and round only at the end; areas and sides here are given to two decimal places, angles to hundredths of a degree.

Worked examples

Example 1

An exact area

Find the area of triangle ABC with a = 6, b = 7 and C = 60^\circ.

Solution

The angle C is between sides a and b, so it is the included angle. With \displaystyle \sin 60^\circ = \dfrac{\sqrt{3}}{2},

\text{Area} = \frac{1}{2}(6)(7)\cdot\frac{\sqrt{3}}{2} = \frac{21\sqrt{3}}{2} \approx 18.19.

The height to side a is 7\sin 60^\circ \approx 6.06, and half of 6 times that is the same number.

Example 2

An obtuse included angle

Danijela, a glazier, cuts a triangular pane of glass with edges of 23 inches and 16 inches meeting at an angle of 141^\circ. What is its area?

Solution

The angle is between the two given edges, so the formula applies directly, obtuse angle and all:

\text{Area} = \frac{1}{2}(23)(16)\sin 141^\circ = 184\sin 141^\circ \approx 115.79.

The pane covers about 115.79 square inches. The sine of 141^\circ is about 0.6293, the same as \sin 39^\circ: the height from the far end of the 16-inch edge, measured to the extended 23-inch edge, is 16\sin 39^\circ.

Example 3

Finding the included angle first

In triangle ABC, A = 35^\circ, B = 80^\circ and c = 14. Find the area.

Solution

Only one side is known, so first find a second. The angle sum gives C = 180^\circ - 35^\circ - 80^\circ = 65^\circ, which pairs c = 14 with its opposite angle. The law of sines gives

a = \frac{14\sin 35^\circ}{\sin 65^\circ} \approx 8.86.

Sides a and c meet at B, so the included angle is B = 80^\circ:

\begin{aligned} \text{Area} &= \frac{1}{2}ac\sin B \\ &= \frac{1}{2}\cdot\frac{14\sin 35^\circ}{\sin 65^\circ}\cdot 14\sin 80^\circ \\ &\approx 61.08. \end{aligned}

Using the full value of a rather than 8.86 keeps the last digit honest. The angle A is not between a and c, and \tfrac{1}{2}ac\sin A would be wrong.

Example 4

Two triangles, one area

50° 130° 13 9 13 9 h h
Figure 3Example 4. Angles of 50^\circ and 130^\circ between the same two sides give the same height, so the same area, but different triangles.

Ljubomir cuts two triangular panes. Each has edges of 9 inches and 13 inches; in one they meet at 50^\circ, in the other at 130^\circ. Compare their areas and their third edges.

Solution

The areas are

\tfrac{1}{2}(9)(13)\sin 50^\circ \quad\text{and}\quad \tfrac{1}{2}(9)(13)\sin 130^\circ.

Since 130^\circ = 180^\circ - 50^\circ, the sines are equal, and both areas are 58.5\sin 50^\circ \approx 44.81 square inches. On a 13-inch base, both heights are 9\sin 50^\circ \approx 6.89 inches.

The third edges come from the law of cosines, and there the sign of the cosine matters:

\begin{aligned} c_1^2 &= 9^2 + 13^2 - 2(9)(13)\cos 50^\circ, \\ c_2^2 &= 9^2 + 13^2 - 2(9)(13)\cos 130^\circ, \end{aligned}

so c_1 \approx 9.98 inches and c_2 \approx 20.01 inches. Same area, different triangles.

Example 5

An angle from an area

Marija wants a triangular pane with edges of 10 inches and 14 inches and an area of 49 square inches. What angle should the edges make, and how long is the third edge?

Solution

Solve the area formula for the sine:

\sin C = \frac{2(49)}{(10)(14)} = \frac{98}{140} = 0.7.

The inverse sine gives C_1 \approx 44.43^\circ, and its supplement C_2 \approx 135.57^\circ has the same sine. Both are possible angles between two edges, so two panes fit.

For the third edge, \displaystyle \cos C = \pm\sqrt{1 - 0.7^2} = \pm\sqrt{0.51}, positive for the acute angle and negative for the obtuse one. The law of cosines gives \displaystyle c^2 = 100 + 196 \mp 280\sqrt{0.51}, so c \approx 9.80 inches when C \approx 44.43^\circ and c \approx 22.27 inches when C \approx 135.57^\circ.

Check: \tfrac{1}{2}(10)(14)(0.7) = 49.

Example 6

A segment at a new angle

\displaystyle \frac{2\pi}{5}
 O A B 12
Figure 4Example 6. The shaded segment is the sector OAB with the triangle OAB taken away.

A round window has radius 12 inches. A straight bar joins two points on its rim whose radii make a central angle of \dfrac{2\pi}{5}. Find the area of the glass between the bar and the rim, exactly and to two decimal places.

Solution

The region is a segment: the sector OAB minus the triangle OAB, as in Arc length and sector area in radians. There the triangle’s area was easy only at \dfrac{\pi}{3} and \dfrac{\pi}{2}. Now it is two sides and the angle between them. The sector has area

\frac{1}{2}(12)^2\cdot\frac{2\pi}{5} = \frac{144\pi}{5}.

The triangle has two sides of 12 meeting at \dfrac{2\pi}{5}, which is 72^\circ, so its area is \tfrac{1}{2}(12)(12)\sin 72^\circ = 72\sin 72^\circ. The segment is

\begin{aligned} &\frac{144\pi}{5} - 72\sin\frac{2\pi}{5} \\ &\approx 22.00 \text{ square inches}. \end{aligned}

In general, a segment of a circle of radius r with central angle \theta in radians has area

\begin{aligned} &\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta \\ &= \frac{1}{2}r^2(\theta - \sin\theta). \end{aligned}
Example 7

A regular octagon

A regular octagon window has its eight corners on a circle of radius 9 inches. Find its area, and compare it with the circle’s.

Solution

Join the center to every corner. That cuts the octagon into 8 congruent triangles, each with two sides of 9 and the angle between them equal to the central angle \dfrac{360^\circ}{8} = 45^\circ. Each has area \tfrac{1}{2}(9)(9)\sin 45^\circ, so

\begin{aligned} \text{Area} &= 8\cdot\frac{81}{2}\cdot\frac{\sqrt{2}}{2} = 162\sqrt{2} \\ &\approx 229.10 \text{ square inches}. \end{aligned}

The circle has area 81\pi \approx 254.47, so the octagon fills about nine tenths of it. The same cut gives any regular polygon with n sides and corners on a circle of radius r the area \dfrac{n}{2}r^2\sin\dfrac{360^\circ}{n}, a route around the apothem used in Area of a regular polygon.

Boundaries

The angle must be the included angle. The formula multiplies two sides by the sine of the angle at the vertex where they meet. With two sides and an angle not between them (SSA), first find the included angle, and the ambiguous case may give two triangles and two areas.

Angles strictly between 0^\circ and 180^\circ. At 0^\circ or 180^\circ the sine is 0: the three points lie on one line and enclose no area.

Going back from an area to an angle gives two answers, one, or none, depending on whether \dfrac{2K}{ab} is less than, equal to, or greater than 1.

Decimal answers are approximations, exact only when the sine is exact, as at 30^\circ, 45^\circ, 60^\circ and their supplements.

Where this goes next

With three sides and no angle, the law of cosines gives an angle and this formula gives the area; doing that algebra once for every triangle produces Heron’s formula. Read with two arrows from one point, ab\sin C is the area of the parallelogram they span, which is the length of the cross product.

Exercises

Two sides and the included angle

Find each area, exactly where the sine is exact, otherwise to two decimal places.

  1. a = 5, b = 8, C = 30^\circ.
  2. b = 4, c = 6, A = 120^\circ.
  3. a = 7, c = 10, B = 45^\circ.
  4. a = 13.5, b = 9.2, C = 77^\circ.
  5. In triangle PQR, PQ = 21, PR = 16 and P = 154^\circ.

When the included angle is not given

  1. A = 40^\circ, C = 75^\circ, b = 20.
  2. Sides 7, 9 and 12. Find the cosine of the largest angle, then the area, exactly.
  3. A = 30^\circ, B = 105^\circ, a = 10. Give the area to two decimal places, and exactly, using the sum formula for sine for \sin 105^\circ = \sin(60^\circ + 45^\circ).

Supplementary angles and working backward

  1. Two triangles have sides 6 and 10, with included angles 35^\circ and 145^\circ. Find their common area and each third side.
  2. A triangle has sides 12 and 15 and area 45. Find every possible included angle.
  3. A triangle has sides 8 and 11 and area 30. Find every possible included angle.
  4. A triangle has two sides of 10 and area 50. Find the included angle.
  5. Can a triangle with sides 6 and 9 have area 30? What is the largest area it can have?
  6. A triangle has area 120, and the angle between a side of 15 and a side x is 64^\circ. Find x.
  7. Danijela cuts a pane with edges of 30 centimeters and 42 centimeters meeting at 58^\circ. The glass costs 0.06 dollars per square centimeter. Find the area and the cost of the glass.

Segments and regular polygons

  1. Find the area of the segment cut from a circle of radius 10 by a chord with central angle \dfrac{5\pi}{6}, exactly and to two decimal places.
  2. Find the area of the segment cut from a circle of radius 20 by a chord with central angle 1 radian.
  3. Find the area of a regular pentagon with corners on a circle of radius 6.
  4. Find the area of a regular hexagon with corners on a circle of radius 4, exactly, and check it by cutting the hexagon into six equilateral triangles.
  5. A chord cuts a circle of radius 7 with a central angle of 100^\circ. Find the areas of the minor and the major segments.

Reasoning

  1. Derive \text{Area} = \tfrac{1}{2}bc\sin A by taking AB as the base and dropping the height from C. Treat an acute A and an obtuse A.
  2. For fixed sides a and b, explain why the area is largest when C = 90^\circ, and describe the area as C moves toward 0^\circ or 180^\circ.
  3. Use the formula to show that an equilateral triangle with side s has area \displaystyle \dfrac{\sqrt{3}}{4}s^2.
  4. Use the law of sines to show that the area of triangle ABC is \dfrac{a^2\sin B\sin C}{2\sin A}.
  5. Given the area and two sides, there may be two triangles. Given the area, one side and the included angle, explain why there is exactly one.
Answers
  1. 10. \tfrac{1}{2}(5)(8)\cdot\tfrac{1}{2} = 10.
  2. \displaystyle 6\sqrt{3} \approx 10.39. \displaystyle \tfrac{1}{2}(4)(6)\sin 120^\circ = 12\cdot\dfrac{\sqrt{3}}{2}.
  3. \displaystyle \dfrac{35\sqrt{2}}{2} \approx 24.75. \displaystyle \tfrac{1}{2}(7)(10)\cdot\dfrac{\sqrt{2}}{2}.
  4. About 60.51. \tfrac{1}{2}(13.5)(9.2)\sin 77^\circ.
  5. About 73.65. Sides PQ and PR meet at P: \tfrac{1}{2}(21)(16)\sin 154^\circ = 168\sin 154^\circ.
  6. About 137.01. B = 65^\circ; a = \dfrac{20\sin 40^\circ}{\sin 65^\circ} \approx 14.18; sides a and b meet at C, so the area is \tfrac{1}{2}ab\sin 75^\circ.
  7. \cos = -\dfrac{1}{9}, area \displaystyle 14\sqrt{5} \approx 31.30. The largest angle is across from 12: \cos C = \dfrac{49 + 81 - 144}{126} = -\dfrac{1}{9}. Then \displaystyle \sin C = \sqrt{1 - \tfrac{1}{81}} = \dfrac{4\sqrt{5}}{9} (positive for any triangle angle), and \displaystyle \tfrac{1}{2}(7)(9)\cdot\dfrac{4\sqrt{5}}{9} = 14\sqrt{5}.
  8. About 68.30, exactly \displaystyle 25(\sqrt{3} + 1). C = 45^\circ; b = \dfrac{10\sin 105^\circ}{\sin 30^\circ} = 20\sin 105^\circ; the area is \tfrac{1}{2}ab\sin C = 100\sin 105^\circ\sin 45^\circ, and the sum formula gives \sin 105^\circ as \sin 60^\circ\cos 45^\circ + \cos 60^\circ\sin 45^\circ, which is \displaystyle \dfrac{\sqrt{6} + \sqrt{2}}{4}.
  9. Area about 17.21; third sides about 6.14 and 15.31. Area 30\sin 35^\circ = 30\sin 145^\circ. Third sides from c^2 = 136 - 120\cos 35^\circ and c^2 = 136 + 120\cos 35^\circ.
  10. 30^\circ or 150^\circ. \sin C = \dfrac{90}{180} = \dfrac{1}{2}.
  11. About 42.99^\circ or 137.01^\circ. \sin C = \dfrac{60}{88} = \dfrac{15}{22}.
  12. 90^\circ only. \sin C = \dfrac{100}{100} = 1.
  13. No. \sin C = \dfrac{60}{54} > 1. The largest area is \tfrac{1}{2}(6)(9) = 27, at a right angle.
  14. About 17.80. x = \dfrac{2(120)}{15\sin 64^\circ}.
  15. About 534.27 square centimeters and 32.06 dollars. \tfrac{1}{2}(30)(42)\sin 58^\circ = 630\sin 58^\circ, times 0.06.
  16. \dfrac{125\pi}{3} - 25 \approx 105.90. Sector \tfrac{1}{2}(100)\cdot\dfrac{5\pi}{6}; triangle \tfrac{1}{2}(100)\sin\dfrac{5\pi}{6} = 25.
  17. About 31.71. \tfrac{1}{2}(400)(1 - \sin 1) = 200(1 - \sin 1), with \sin 1 taken in radians.
  18. About 85.60. Five triangles with central angle 72^\circ: 5\cdot\tfrac{1}{2}(36)\sin 72^\circ = 90\sin 72^\circ.
  19. \displaystyle 24\sqrt{3} \approx 41.57. Six triangles with central angle 60^\circ: \displaystyle 6\cdot\tfrac{1}{2}(16)\cdot\dfrac{\sqrt{3}}{2}. Each triangle has two sides of 4 and a 60^\circ angle between them, so it is equilateral with area \displaystyle \dfrac{\sqrt{3}}{4}(16) = 4\sqrt{3}, and six of them make \displaystyle 24\sqrt{3}.
  20. Minor about 18.63, major about 135.31. 100^\circ = \dfrac{5\pi}{9}, so the minor segment is \tfrac{1}{2}(49)\left(\dfrac{5\pi}{9} - \sin 100^\circ\right), and the major one is 49\pi minus that.
  21. With AB = c as the base, the height h from C meets line AB at D. If A is acute and D is on the side AB or past B, right triangle ADC has hypotenuse b and angle A, so h = b\sin A. If A is obtuse, D lies on AB extended past A, the right triangle holds 180^\circ - A, and h = b\sin(180^\circ - A) = b\sin A. Either way the area is \tfrac{1}{2}c\,(b\sin A).
  22. The area \tfrac{1}{2}ab\sin C is a fixed positive number times \sin C, and for 0^\circ < C < 180^\circ the sine is largest, 1, at 90^\circ. As C moves toward 0^\circ or 180^\circ, \sin C moves toward 0, the height shrinks, and the triangle flattens toward a segment with area near 0.
  23. All angles are 60^\circ, so the area is \displaystyle \tfrac{1}{2}s\cdot s\sin 60^\circ = \tfrac{1}{2}s^2\cdot\dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{4}s^2.
  24. Start from \tfrac{1}{2}ab\sin C. The law of sines gives b = \dfrac{a\sin B}{\sin A}. Substituting, the area is \tfrac{1}{2}a\cdot\dfrac{a\sin B}{\sin A}\cdot\sin C = \dfrac{a^2\sin B\sin C}{2\sin A}.
  25. Solving K = \tfrac{1}{2}ab\sin C for b gives b = \dfrac{2K}{a\sin C}, one number, since \sin C > 0 is known and dividing by it is allowed. With a, b and C fixed, SAS fixes the triangle. Solving for the angle instead asks which angles have a given sine, and C and 180^\circ - C share every sine below 1.
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