Precalculus by Richard Wright

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6-02 Law of Cosines

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.6.3

swamp
Figure 1: Swamp. credit (pixnio.com/USFWS modified by RW)

To approximate the length of an impassible swamp, a surveyor finds a point midway along the side of the swamp but at a distance where he can measure the distance to each end. He measures one distance as 235 m and the other as 290 m. The angle between the distance lines is 110°. The law of sines cannot be used to find the third side because a side and opposite angle are not known. So the surveyor will have to use the law of cosines to find the length of the swamp.

Law of Cosines

To solve a triangle, a side and two other pieces of information must be known. The law of cosines should be used when two sides and the included angle (SAS) or all three sides (SSS) are known.

Law of Cosines

Use for SAS and SSS cases

$$a^2 = b^2 + c^2 - 2bc \cos A$$

$$b^2 = a^2 + c^2 - 2ac \cos B$$

$$c^2 = a^2 + b^2 - 2ab \cos C$$

Notice that the first and last letter are the same—one a side and one an angle. The other two sides are in between.

Example 1: Solve a SAS Triangle with the Law of Cosines

Solve ∆ABC where A = 30°, b = 15, and c = 17.

Solution
Figure 2

Because angle A is known, start by finding side a.

$$a^2 = b^2 + c^2 - 2bc \cos A$$

$$a^2 = 15^2 + 17^2 - 2\left(15\right)\left(17\right) \cos 30°$$

$$a^2 ≈ 72.32704$$

a ≈ 8.50

Now, find an angle.

$$b^2 = a^2 + c^2 - 2ac \cos B$$

$$15^2 = 8.50^2 + 17^2 - 2\left(8.50\right)\left(17\right) \cos B$$

$$225 = 72.25 + 289 - 289 \cos B$$

Subtract the 72.25 and 289.

$$-136.25 = -289 \cos B$$

$$0.4715 = \cos B$$

$$B = \cos^{-1} 0.4715$$

B ≈ 61.87°

Lastly, subtract from 180° to find the last angle.

C = 180° – AB

C ≈ 180° – 30° – 61.87°

C ≈ 88.13°

Try It 1

Solve ΔABC where B = 120°, a = 20, and c = 18.

Answer

A = 31.74°, C = 28.26°, b = 32.92

Notice in the previous example, that after finding the first side, a side and opposite angle is known. You might think that you could use the law of sines to find angle B, but that is a bad idea. If you used the law of sines, it would be the SSA case where there is a possibility of two solutions. However, because we started with SAS or SSS for law of cosines, there is only one solution since they are both geometry congruence theorems. If you start a problem with the law of cosines, you need to continue using the law of cosines. Do not switch to the law of sines!

Example 2: Solve a SSS Triangle with the Law of Cosines

Solve ∆RST where r = 20, s = 17, t = 24.

Solution
Figure 3

Start by finding an angle.

$$r^2 = s^2 + t^2 - 2st \cos R$$

$$20^2 = 17^2 + 24^2 - 2\left(17\right)\left(24\right) \cos R$$

$$400 = 289 + 576 - 816 \cos R$$

$$-465 = -816 \cos R$$

$$\frac{155}{272} = \cos R$$

$$R = \cos^{-1} \frac{155}{272}$$

R ≈ 55.26°

Now find a second angle.

$$s^2 = r^2 + t^2 - 2rt \cos S$$

$$17^2 = 20^2 + 24^2 - 2\left(20\right)\left(24\right) \cos S$$

$$289 = 400 + 576 - 960 \cos S$$

$$-687 = -960 \cos S$$

$$\frac{229}{320} = \cos S$$

$$S = \cos^{-1} \frac{229}{320}$$

S ≈ 44.31°

Subtract to find the last angle.

T = 180° – RS

T ≈ 180° – 55.26° – 44.31°

T ≈ 80.43°

Try It 2

Solve ΔABC where a = 20, b = 22, and c = 18.

Answer

A = 58.99°, B = 70.53°, C = 50.48°

Area of a Triangle: Heron’s Formula

In the last lesson, the area of a triangle was found using two sides and an angle. The Greek mathematician, Heron, developed a formula for finding the area of a triangle when just the lengths of the sides are known.

Heron’s Area Formula

$$Area = \sqrt{s\left(s - a\right)\left(s - b\right)\left(s - c\right)}$$

where \(s = \frac{a + b + c}{2}\)

Example 3: Find the Area of a Triangle with Known Sides

Find the area of a triangle with side lengths 10, 12, 15.

Solution

Start by finding s.

$$s = \frac{a + b + c}{2}$$

$$s = \frac{10 + 12 + 15}{2}$$

$$s = \frac{37}{2}$$

Now use Heron’s formula

$$Area = \sqrt{\frac{37}{2}\left(\frac{37}{2} - 10\right)\left(\frac{37}{2} - 12\right)\left(\frac{37}{2} - 15\right)}$$

$$Area ≈ 59.81$$

Try It 3

Find the area of ΔABC where a = 20, b = 22, and c = 18.

Answer

169.71

Lesson Summary
Law of Cosines

Use for SAS and SSS cases

$$a^2 = b^2 + c^2 - 2bc \cos A$$

$$b^2 = a^2 + c^2 - 2ac \cos B$$

$$c^2 = a^2 + b^2 - 2ab \cos C$$

Heron’s Area Formula

$$Area = \sqrt{s\left(s - a\right)\left(s - b\right)\left(s - c\right)}$$

where \(s = \frac{a + b + c}{2}\)

Helpful videos about this lesson.

Practice Exercises (*optional)

  1. When should you use the law of sines? Law of cosines? The simple sine, cosine, and tangent ratios?
  2. Solve the given triangles.

  3. ∆ABC where A = 109°, b = 31, and c = 28
  4. ∆DFG where d = 102, f = 96, and g = 57
  5. ∆HJK where h = 18, j = 15, and k = 28
  6. *∆LMN where L = 61°, m = 7, and n = 13
  7. Solve the given triangles using the law of sines or law of cosines.

  8. ∆PQR where R = 34°, p = 43, and q = 51
  9. ∆STU where T = 124°, t = 63, and u = 36
  10. Find the area of the triangle.

  11. ∆DFG where d = 102, f = 96, and g = 57
  12. ∆HJK where h = 18, j = 15, and k = 28
  13. ∆WXY where w = 3, x = 4, y = 6
  14. Problem Solving

  15. To approximate the length of an impassible swamp, a surveyor finds a point midway along the side of the swamp but at a distance where he can measure the distance to each end. He measures one distance as 235 m and the other as 290 m. The angle between the distance lines is 110°. How far across is the swamp?
    swamp
  16. Mixed Review

  17. (6-01) Find the area of ∆MXD where M = 78°, x = 20, d = 24.
  18. (6-01) Solve triangle ∆CUP where C = 63°, c = 68, and p = 71.
  19. (4-10) Solve the right triangle.
  20. (4-01) For the angle 165° a) draw the angle in standard position, b) convert it to the other angle unit, c) find a positive coterminal angle, d) find a negative coterminal angle, e) find the complementary angle, and f) find the supplementary angle.
  21. (3-03) Condense \(\log_3 x + \log_3 \left(x+1\right) - \log_3 y\).

Answers

  1. Law of Sines: AAS, ASA, SSA; Law of Cosines: SSS, SAS; Tangent Ratios: Right triangles
  2. a = 48.06, B = 37.58°, C = 33.42°
  3. D = 79.15°, F = 67.57°, G = 33.29°
  4. H = 35.37°, J = 28.84°, K = 115.80°
  5. = 11.39, M = 32.51°, N = 86.49°
  6. r = 28.53, P = 57.44°, Q = 88.56°
  7. s = 35.35, S = 27.72°, U = 28.28°
  8. 2687.05
  9. 121.55
  10. 5.33
  11. 431.21 m
  12. 234.76
  13. 1st Solution: u = 57.17, U = 48.52°, P = 68.48°; 2nd Solution: u = 7.29, U = 5.48°, P = 111.52°
  14. a = 8.06, B = 29.75°, C = 60.26°
  15. ; \(\frac{11π}{12}\); 525°; -195°; none; 15°
  16. \(\log_3 \frac{x\left(x+1\right)}{y}\)