Law of Cosines: Solving for a Side

Question 1 of 3

Find

C B

$CB$

Round your answer to

1

$1$ decimal place

$C B =$ $CB=$ (4.8) $c m$ $cm$

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Cosine Law

a^{2} = b^{2} + c^{2} - 2 b c \cos A

$\color{#007DDC}{a}^2=\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$

where:
$a$ $a$ is the side opposite angle $A$ $A$
$b$ $b$ is the side opposite angle $B$ $B$
$c$ $c$ is the side opposite angle $C$ $C$

When to use the Cosine Law (for non-right angled triangles)

a) Given 3 sides to find an angle

or

b) Given 2 sides and 1 angle to find the other side

Since

2

$2$ sides are given together with an angle between them, use the Cosine Law.

First, label the triangle according to the Cosine Law.

Substitute the three known values to the Cosine Law to find the length of side

C B

$CB$ or $a$ $a$ .

From labelling the triangle, we know that the known values are those with labels

A, b

$A, b$ and

c

$c$ .

A = 42 °

$A=42°$

b = 7 c m

$b=7 cm$

c = 6 c m

$c=6 cm$

$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$
$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{7}^2+\color{#9a00c7}{6}^2-2(\color{#00880A}{7})(\color{#9a00c7}{6})\cos\color{#007DDC}{42°}$	Substitute the values
$a^{2}$ $a^2$	$=$ $=$	$49 + 36 - 84 \cos 42 °$ $49+36-84cos42°$	Simplify
$a^{2}$ $a^2$	$=$ $=$	$85 - 84$ $85-84$ $\cos 42 °$ $cos42°$	Evaluate $\cos$ $cos$ $42$ $42$ on your calculator
$a^{2}$ $a^2$	$=$ $=$	$85 - 84$ $85-84$ $(0.7431448)$ $(0.7431448)$	Simplify
$a^{2}$ $a^2$	$=$ $=$	$85 - 62.42416$ $85-62.42416$
$a^{2}$ $a^2$	$=$ $=$	$22.57583$ $22.57583$
$\sqrt{a^{2}}$ $sqrt(a^2)$	$=$ $=$	$\sqrt{22.57583}$ $sqrt22.57583$	Take the square root of both sides
$a$ $a$	$=$ $=$	$4.75$ $4.75$
$a$ $a$ or $C B$ $CB$	$=$ $=$	$4.8 c m$ $4.8 cm$	Round off to $1$ $1$ decimal place

4.8 c m

$4.8 cm$

Question 2 of 3

Find

A C

$AC$

Round your answer to

1

$1$ decimal place

$A C =$ $AC=$ (17.2) $m$ $m$

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Cosine Law

a^{2} = b^{2} + c^{2} - 2 b c \cos A

$\color{#007DDC}{a}^2=\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$

where:
$a$ $a$ is the side opposite angle $A$ $A$
$b$ $b$ is the side opposite angle $B$ $B$
$c$ $c$ is the side opposite angle $C$ $C$

When to use the Cosine Law (for non-right angled triangles)

a) Given 3 sides to find an angle

or

b) Given 2 sides and 1 angle to find the other side

Since

2

$2$ sides are given together with an angle between them, use the Cosine Law.

First, label the triangle according to the Cosine Law.

Substitute the three known values to the Cosine Law to find the length of side

A C

$AC$ or $b$ $b$ .

From labelling the triangle, we know that the known values are those with labels

B, a

$B, a$ and

c

$c$ .

B = 117 °

$B=117°$

a = 12 m

$a=12 m$

c = 8 m

$c=8 m$

$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$
$\color{#00880A}{b}^2$	$=$ $=$	$\color{#007DDC}{a}^2+\color{#9a00c7}{c}^2-2\color{#007DDC}{a}\color{#9a00c7}{c}\cos\color{#00880A}{B}$	Rewrite formula according to given values
$\color{#00880A}{b}^2$	$=$ $=$	$\color{#007DDC}{12}^2+\color{#9a00c7}{8}^2-2(\color{#007DDC}{12})(\color{#9a00c7}{8})\cos\color{#00880A}{117°}$	Substitute the values
$b^{2}$ $b^2$	$=$ $=$	$144 + 64 - 192$ $144+64-192$ $\cos 117 °$ $cos117°$	Evaluate $\cos$ $cos$ $117$ $117$ on your calculator
$b^{2}$ $b^2$	$=$ $=$	$208 - 192$ $208-192$ $(- 0.45399)$ $(-0.45399)$
$b^{2}$ $b^2$	$=$ $=$	$208 + 87.16676$ $208+87.16676$
$b^{2}$ $b^2$	$=$ $=$	$295.166$ $295.166$
$\sqrt{b^{2}}$ $sqrt(b^2)$	$=$ $=$	$\sqrt{295.166}$ $sqrt295.166$	Take the square root of both sides
$b$ $b$	$=$ $=$	$17.18$ $17.18$
$b$ $b$ or $A C$ $AC$	$=$ $=$	$17.2 m$ $17.2 m$	Round off to $1$ $1$ decimal place

17.2 m

$17.2 m$

Question 3 of 3

Find

A B

$AB$

Round your answer to

1

$1$ decimal place

$A B =$ $AB=$ (79.1) $m$ $m$

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Cosine Law

a^{2} = b^{2} + c^{2} - 2 b c \cos A

$\color{#007DDC}{a}^2=\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$

where:
$a$ $a$ is the side opposite angle $A$ $A$
$b$ $b$ is the side opposite angle $B$ $B$
$c$ $c$ is the side opposite angle $C$ $C$

When to use the Cosine Law (for non-right angled triangles)

a) Given 3 sides to find an angle

or

b) Given 2 sides and 1 angle to find the other side

Since

2

$2$ sides are given together with an angle between them, use the Cosine Law.

First, label the triangle according to the Cosine Law.

Substitute the three known values to the Cosine Law to find the length of side

A B

$AB$ or $c$ $c$ .

From labelling the triangle, we know that the known values are those with labels

C, a

$C, a$ and

b

$b$ .

C = 139 °

$C=139°$

a = 31 m

$a=31 m$

b = 53 m

$b=53 m$

$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$
$\color{#9a00c7}{c}^2$	$=$ $=$	$\color{#007DDC}{a}^2+\color{#00880A}{b}^2-2\color{#007DDC}{a}\color{#00880A}{b}\cos\color{#9a00c7}{C}$	Rewrite formula according to given values
$\color{#9a00c7}{c}^2$	$=$ $=$	$\color{#007DDC}{31}^2+\color{#00880A}{53}^2-2\color{#007DDC}{31}\color{#00880A}{53}\cos\color{#9a00c7}{139°}$	Substitute the values
$c^{2}$ $c^2$	$=$ $=$	$961 + 2809 - 1643$ $961+2809-1643$ $\cos 139 °$ $cos139°$	Evaluate $\cos$ $cos$ $139$ $139$ on your calculator
$c^{2}$ $c^2$	$=$ $=$	$3770 - 1643$ $3770-1643$ $(- 0.7547096)$ $(-0.7547096)$
$c^{2}$ $c^2$	$=$ $=$	$3770 + 2479.97568$ $3770+2479.97568$
$c^{2}$ $c^2$	$=$ $=$	$6249.97568$ $6249.97568$
$\sqrt{c^{2}}$ $sqrt(c^2)$	$=$ $=$	$\sqrt{6249.97568}$ $sqrt6249.97568$	Take the square root of both sides
$c$ $c$	$=$ $=$	$79.05679$ $79.05679$
$c$ $c$ or $A B$ $AB$	$=$ $=$	$79.1 m$ $79.1 m$	Round off to $1$ $1$ decimal place

79.1 m

$79.1 m$

Law of Cosines: Solving for a Side

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Quiz summary

Information

1. Question

Hint

Correct!

Cosine Law

When to use the Cosine Law (for non-right angled triangles)

2. Question

Hint

Great Work!

Cosine Law

When to use the Cosine Law (for non-right angled triangles)

3. Question

Hint

Fantastic!

Cosine Law

When to use the Cosine Law (for non-right angled triangles)

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