LAW OF COSINE

Derivation

From the right triangle to the left

cos 𝐴 = 𝑥

𝑐 ; 𝑥 = 𝑐 cos 𝐴 (1)

𝑐2 = ℎ2 + 𝑥2 ; ℎ2 = 𝑐2 − 𝑥2 (2)

From the right triangle to the right

𝑎2 = ℎ2 + (𝑏 − 𝑥)2

𝑎2 = ℎ2 + 𝑏2 − 2𝑏𝑥 + 𝑥2 (3) Substitute equation (2) to equation (3)

𝑎2 = 𝑐2 − 𝑥2 + 𝑏2 − 2𝑏𝑥 + 𝑥2

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑥 (4) Substitute equation (1) to (4)

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴 Therefore 𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

𝑏2 = 𝑎2 + 𝑐2 − 2𝑎𝑐 cos 𝐵

𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶

b

c

A C

a

h

(x,y)

(a,0) x b-x

B

CASE III: Two side and the included angle given Problem 1

Solve the unknown side and angles of a triangle for which a = 25, b = 30 and C = 50⁰.

Find

c = ? A = ? B = ?

Solution

𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶

𝑐2 = 252 + 302 − 2(25)(30) cos 50⁰

𝑐2 = 560.82

∴ 𝑐 = 23.68 Using sine law

𝑎

sin 𝐴=

𝑏

sin 𝐵

25

sin 𝐴=

23.68

sin 50⁰

sin 𝐴 = (25) sin 50⁰

23.68

sin 𝐴 = 0.8087 …

𝐴 = sin−1(0.8087 …)

𝐴1 = 53.970

𝐴2 = 1800 − 𝐴1 = 1800 − 53.970 = 126.03⁰

Since the side 𝑎 is the second lowest side, therefore its opposite angle must be acute.

∴ 𝐴 = 53.97⁰

Using sine law

𝑏

sin 𝐵=

𝑏

sin 𝐵

30

sin 𝐵=

23.68

sin 50⁰

sin 𝐵 = (30) sin 50⁰

23.68

sin 𝐵 = 0.9705 …

𝐵 = sin−1(09705 …)

𝐵1 = 76.05⁰

𝐵2 = 1800 − 𝐵1 = 1800 − 76.050 = 103.95⁰

Since side 𝑏 is the largest side, adding the acute angle 𝐵1 to the angles 𝐴 and 𝐶 will give 180⁰.

∴ 𝐵 = 76.050

Checking 𝐴 + 𝐵 + 𝐶 = 180⁰

53.970 + 76.050 + 500 = 1800 (𝑂𝐾)

CASE IV: Three sides given Problem 2

Solve the unknown angles of a triangle for which a = 5, b = 6 and c = 9.

Find

A = ? B = ? C = ?

Solution

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

2𝑏𝑐 cos 𝐴 = 𝑏2 + 𝑐2 − 𝑎2

cos 𝐴 = 𝑏2 + 𝑐2 − 𝑎2

2𝑏𝑐

cos 𝐴 = 62 + 92 − 52

2(6)(9)

cos 𝐴 = 0.85185 …

𝐴 = cos−1(0.85185 … )

∴ 𝐴 = 31.59⁰

Using Cosine Law (Sine Law can also be used)

𝑏2 = 𝑎2 + 𝑐2 − 2𝑎𝑐 cos 𝐵

2𝑎𝑐 cos 𝐵 = 𝑎2 + 𝑐2 − 𝑏2

cos 𝐵 = 𝑎2 + 𝑐2 − 𝑏2

2𝑎𝑐

cos 𝐵 = 52 + 92 − 62

2(5)(9)

cos 𝐵 = 0.7777 …

𝐵 = cos−1(0.7777 … )

∴ 𝐵 = 38.94⁰

Using Cosine Law (Sine Law can also be used)

𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶

2𝑎𝑏 cos 𝐶 = 𝑎2 + 𝑏2 − 𝑐2

cos 𝐶 = 𝑎2 + 𝑏2 − 𝑐2

2𝑎𝑏

cos 𝐶 = 52 + 62 − 92

2(5)(6)

cos 𝐶 = 0.3333 …

𝐶 = cos−1(0.3333 … )

∴ 𝐶 = 109.47⁰

Checking 𝐴 + 𝐵 + 𝐶 = 180⁰

31.590 + 38.940 + 109.470 = 1800 (𝑂𝐾)

CASE II: Given two side and one opposite angle (Possible triangle/s can be formed) A. Two Possible Triangles

Problem 3

Solve for the possible side/s c of a triangle for which: a = 25.2 A = 54.2⁰ b = 30.5 Solution: Using Cosine Law

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

25.22 = 30.52 + 𝑐2 − 2(30.5)𝑐 cos 54.2⁰

0 = 30.52 − 25.22 + 𝑐2 − 2(30.5)𝑐 cos 54.2⁰

𝑐2 − 35.68𝑐 + 295.21 = 0

Using quadratic formula

𝑥 =−𝐵 ± √𝐵2 − 4𝐴𝐶

2𝐴

𝐴 = 1 𝐵 = −35.68 𝐶 = 295.21

𝑐1 =−(−35.68) + √(−35.68)2 − 4(1)(295.21)

2(1)

∴ 𝑐1 = +22.64

𝑐2 =−(−35.68) − √(−35.68)2 − 4(1)(295.21)

2(1)

∴ 𝑐2 = +13.04

Since there are two positive real roots, thus there are two possible sides of c. Therefore, two triangles can be formed. Use cosine law (or sine law) to solve for the possible angles B & C for side c = 22.64 and possible angles B & C for side c = 13.04.

B. One Possible Triangle

Problem 4

Solve for the possible side/s c of a triangle for which: a = 5.21 A = 47.6⁰ b = 3.06 Solution: Using Cosine Law

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

5.212 = 3.062 + 𝑐2 − 2(3.06)𝑐 cos 47.6⁰

0 = 3.062 − 5.212 + 𝑐2 − 2(3.06)𝑐 cos 47.6⁰

𝑐2 − 4.13𝑐 − 17.78 = 0

Using quadratic formula

𝑥 =−𝐵 ± √𝐵2 − 4𝐴𝐶

2𝐴

𝐴 = 1 𝐵 = −4.13 𝐶 = −17.78

𝑐1 =−(−4.13) + √(−4.13)2 − 4(1)(−17.78)

2(1)

∴ 𝑐1 = +6.76

𝑐2 =−(−4.13) − √(−4.13)2 − 4(1)(−17.78)

2(1)

𝑐2 = −2.63

Since there is only one positive real root, thus there is only one possible side of c. Therefore, one triangle can be formed. Use cosine law/sine law to solve for the possible angles B and C for side c = 6.67.

C. No Possible Triangle Problem 5

Solve for the possible side/s c of a triangle for which: a = 2.30 A = 42⁰ b = 4.50 Solution: Using Cosine Law

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

2.32 = 4.52 + 𝑐2 − 2(4.5)𝑐 cos 42⁰

0 = 4.52 − 2.32 + 𝑐2 − 2(4.5)𝑐 cos 42⁰

𝑐2 − 6.69𝑐 + 14.96 = 0

Using quadratic formula

𝑥 =−𝐵 ± √𝐵2 − 4𝐴𝐶

2𝐴

𝐴 = 1 𝐵 = −6.69 𝐶 = +14.96

𝑐1 =−(−6.69) + √(−6.69)2 − 4(1)(+14.96)

2(1)

𝑐1 =6.69 + √−15.08

2 =

6.69 + √15.08(−1)

2 =

6.69 + 3.87𝑖

2

𝑐1 = 3.35 + 1.97𝑖 (𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦)

𝑐2 =−(−6.69) − √(−6.69)2 − 4(1)(+14.96)

2(1)

𝑐2 =6.69 − √−15.08

2 =

6.69 − √15.08(−1)

2 =

6.69 − 3.87𝑖

2

𝑐2 = 3.35 − 1.97𝑖 (𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦)

Since there is no positive real root, thus there is no possible side of c. Therefore, no triangle can be formed.