Trigonometry Examples



\begin{matrix} Side & Angle \\ \begin{matrix} b = & 28 \\ c = & 53 \\ a = & 45 \end{matrix} & \begin{matrix} A = \\ B = \\ C = \end{matrix} \end{matrix}

$\begin{matrix} Side & Angle \\ \begin{matrix} b = & 28 \\ c = & 53 \\ a = & 45 \end{matrix} & \begin{matrix} A = \\ B = \\ C = \end{matrix} \end{matrix}$

Step 1

Use the law of cosines to find the unknown side of the triangle, given the other two sides and the included angle.

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

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

Step 2

Solve the equation.

A = \arccos (\frac{b^{2} + c^{2} - a^{2}}{2 b c})

$A = \arccos (\frac{b^{2} + c^{2} - a^{2}}{2 b c})$

Step 3

Substitute the known values into the equation.

A = \arccos (\frac{{(28)}^{2} + {(53)}^{2} - {(45)}^{2}}{2 (28) (53)})

$A = \arccos (\frac{{(28)}^{2} + {(53)}^{2} - {(45)}^{2}}{2 (28) (53)})$

Step 4

Simplify the results.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise

28

$28$ to the power of

2

$2$ .

A = \arccos (\frac{784 + 53^{2} - 45^{2}}{2 (28) \cdot 53})

$A = \arccos (\frac{784 + 53^{2} - 45^{2}}{2 (28) \cdot 53})$

Step 4.1.2

Raise

53

$53$ to the power of

2

$2$ .

A = \arccos (\frac{784 + 2809 - 45^{2}}{2 (28) \cdot 53})

$A = \arccos (\frac{784 + 2809 - 45^{2}}{2 (28) \cdot 53})$

Step 4.1.3

Raise

45

$45$ to the power of

2

$2$ .

A = \arccos (\frac{784 + 2809 - 1 \cdot 2025}{2 (28) \cdot 53})

$A = \arccos (\frac{784 + 2809 - 1 \cdot 2025}{2 (28) \cdot 53})$

Step 4.1.4

Multiply

- 1

$- 1$ by

2025

$2025$ .

A = \arccos (\frac{784 + 2809 - 2025}{2 (28) \cdot 53})

$A = \arccos (\frac{784 + 2809 - 2025}{2 (28) \cdot 53})$

Step 4.1.5

Add

784

$784$ and

2809

$2809$ .

A = \arccos (\frac{3593 - 2025}{2 (28) \cdot 53})

$A = \arccos (\frac{3593 - 2025}{2 (28) \cdot 53})$

Step 4.1.6

Subtract

2025

$2025$ from

3593

$3593$ .

A = \arccos (\frac{1568}{2 (28) \cdot 53})

$A = \arccos (\frac{1568}{2 (28) \cdot 53})$

A = \arccos (\frac{1568}{2 (28) \cdot 53})

$A = \arccos (\frac{1568}{2 (28) \cdot 53})$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply

2

$2$ by

28

$28$ .

A = \arccos (\frac{1568}{56 \cdot 53})

$A = \arccos (\frac{1568}{56 \cdot 53})$

Step 4.2.2

Multiply

56

$56$ by

53

$53$ .

A = \arccos (\frac{1568}{2968})

$A = \arccos (\frac{1568}{2968})$

A = \arccos (\frac{1568}{2968})

$A = \arccos (\frac{1568}{2968})$

Step 4.3

Cancel the common factor of

1568

$1568$ and

2968

$2968$ .

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Step 4.3.1

Factor

56

$56$ out of

1568

$1568$ .

A = \arccos (\frac{56 (28)}{2968})

$A = \arccos (\frac{56 (28)}{2968})$

Step 4.3.2

Cancel the common factors.

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Step 4.3.2.1

Factor

56

$56$ out of

2968

$2968$ .

A = \arccos (\frac{56 \cdot 28}{56 \cdot 53})

$A = \arccos (\frac{56 \cdot 28}{56 \cdot 53})$

Step 4.3.2.2

Cancel the common factor.

A = \arccos (\frac{56 \cdot 28}{56 \cdot 53})

$A = \arccos (\frac{56 \cdot 28}{56 \cdot 53})$

Step 4.3.2.3

Rewrite the expression.

A = \arccos (\frac{28}{53})

$A = \arccos (\frac{28}{53})$

A = \arccos (\frac{28}{53})

$A = \arccos (\frac{28}{53})$

A = \arccos (\frac{28}{53})

$A = \arccos (\frac{28}{53})$

Step 4.4

Evaluate

\arccos (\frac{28}{53})

$\arccos (\frac{28}{53})$ .

A = 58.10920819

$A = 58.10920819$

A = 58.10920819

$A = 58.10920819$

Step 5

Use the law of cosines to find the unknown side of the triangle, given the other two sides and the included angle.

b^{2} = a^{2} + c^{2} - 2 a c \cos (B)

$b^{2} = a^{2} + c^{2} - 2 a c \cos (B)$

Step 6

Solve the equation.

B = \arccos (\frac{a^{2} + c^{2} - b^{2}}{2 a c})

$B = \arccos (\frac{a^{2} + c^{2} - b^{2}}{2 a c})$

Step 7

Substitute the known values into the equation.

B = \arccos (\frac{{(45)}^{2} + {(53)}^{2} - {(28)}^{2}}{2 (45) (53)})

$B = \arccos (\frac{{(45)}^{2} + {(53)}^{2} - {(28)}^{2}}{2 (45) (53)})$

Step 8

Simplify the results.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Raise

45

$45$ to the power of

2

$2$ .

B = \arccos (\frac{2025 + 53^{2} - 28^{2}}{2 (45) \cdot 53})

$B = \arccos (\frac{2025 + 53^{2} - 28^{2}}{2 (45) \cdot 53})$

Step 8.1.2

Raise

53

$53$ to the power of

2

$2$ .

B = \arccos (\frac{2025 + 2809 - 28^{2}}{2 (45) \cdot 53})

$B = \arccos (\frac{2025 + 2809 - 28^{2}}{2 (45) \cdot 53})$

Step 8.1.3

Raise

28

$28$ to the power of

2

$2$ .

B = \arccos (\frac{2025 + 2809 - 1 \cdot 784}{2 (45) \cdot 53})

$B = \arccos (\frac{2025 + 2809 - 1 \cdot 784}{2 (45) \cdot 53})$

Step 8.1.4

Multiply

- 1

$- 1$ by

784

$784$ .

B = \arccos (\frac{2025 + 2809 - 784}{2 (45) \cdot 53})

$B = \arccos (\frac{2025 + 2809 - 784}{2 (45) \cdot 53})$

Step 8.1.5

Add

2025

$2025$ and

2809

$2809$ .

B = \arccos (\frac{4834 - 784}{2 (45) \cdot 53})

$B = \arccos (\frac{4834 - 784}{2 (45) \cdot 53})$

Step 8.1.6

Subtract

784

$784$ from

4834

$4834$ .

B = \arccos (\frac{4050}{2 (45) \cdot 53})

$B = \arccos (\frac{4050}{2 (45) \cdot 53})$

B = \arccos (\frac{4050}{2 (45) \cdot 53})

$B = \arccos (\frac{4050}{2 (45) \cdot 53})$

Step 8.2

Simplify the denominator.

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Step 8.2.1

Multiply

2

$2$ by

45

$45$ .

B = \arccos (\frac{4050}{90 \cdot 53})

$B = \arccos (\frac{4050}{90 \cdot 53})$

Step 8.2.2

Multiply

90

$90$ by

53

$53$ .

B = \arccos (\frac{4050}{4770})

$B = \arccos (\frac{4050}{4770})$

B = \arccos (\frac{4050}{4770})

$B = \arccos (\frac{4050}{4770})$

Step 8.3

Cancel the common factor of

4050

$4050$ and

4770

$4770$ .

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Step 8.3.1

Factor

90

$90$ out of

4050

$4050$ .

B = \arccos (\frac{90 (45)}{4770})

$B = \arccos (\frac{90 (45)}{4770})$

Step 8.3.2

Cancel the common factors.

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Step 8.3.2.1

Factor

90

$90$ out of

4770

$4770$ .

B = \arccos (\frac{90 \cdot 45}{90 \cdot 53})

$B = \arccos (\frac{90 \cdot 45}{90 \cdot 53})$

Step 8.3.2.2

Cancel the common factor.

B = \arccos (\frac{90 \cdot 45}{90 \cdot 53})

$B = \arccos (\frac{90 \cdot 45}{90 \cdot 53})$

Step 8.3.2.3

Rewrite the expression.

B = \arccos (\frac{45}{53})

$B = \arccos (\frac{45}{53})$

B = \arccos (\frac{45}{53})

$B = \arccos (\frac{45}{53})$

B = \arccos (\frac{45}{53})

$B = \arccos (\frac{45}{53})$

Step 8.4

Evaluate

\arccos (\frac{45}{53})

$\arccos (\frac{45}{53})$ .

B = 31.8907918

$B = 31.8907918$

B = 31.8907918

$B = 31.8907918$

Step 9

The sum of all the angles in a triangle is

180

$180$ degrees.

58.10920819 + C + 31.8907918 = 180

$58.10920819 + C + 31.8907918 = 180$

Step 10

Solve the equation for

C

$C$ .

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Step 10.1

Add

58.10920819

$58.10920819$ and

31.8907918

$31.8907918$ .

C + 90 = 180

$C + 90 = 180$

Step 10.2

Move all terms not containing

C

$C$ to the right side of the equation.

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Step 10.2.1

Subtract

90

$90$ from both sides of the equation.

C = 180 - 90

$C = 180 - 90$

Step 10.2.2

Subtract

90

$90$ from

180

$180$ .

C = 90

$C = 90$

C = 90

$C = 90$

C = 90

$C = 90$

Step 11

These are the results for all angles and sides for the given triangle.

A = 58.10920819

$A = 58.10920819$

B = 31.8907918

$B = 31.8907918$

C = 90

$C = 90$

a = 45

$a = 45$

b = 28

$b = 28$

c = 53

$c = 53$