Trigonometry Examples

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\begin{matrix} Side & Angle \begin{matrix} b = & 36 c = & 39 a = & 15 \end{matrix} & \begin{matrix} A = B = C = \end{matrix} \end{matrix}

$\begin{matrix} Side & Angle \\ \begin{matrix} b = & 36 \\ c = & 39 \\ a = & 15 \end{matrix} & \begin{matrix} A = \\ B = \\ C = \end{matrix} \end{matrix}$

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{{(36)}^{2} + {(39)}^{2} - {(15)}^{2}}{2 (36) (39)})

$A = \arccos (\frac{{(36)}^{2} + {(39)}^{2} - {(15)}^{2}}{2 (36) (39)})$

Step 4

Simplify the results.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise

36

$36$ to the power of

2

$2$ .

A = arccos (\frac{1296 + 39^{2} - 15^{2}}{2 (36) \cdot 39})

$A = \arccos (\frac{1296 + 39^{2} - 15^{2}}{2 (36) \cdot 39})$

Step 4.1.2

Raise

39

$39$ to the power of

2

$2$ .

A = arccos (\frac{1296 + 1521 - 15^{2}}{2 (36) \cdot 39})

$A = \arccos (\frac{1296 + 1521 - 15^{2}}{2 (36) \cdot 39})$

Step 4.1.3

Raise

15

$15$ to the power of

2

$2$ .

A = arccos (\frac{1296 + 1521 - 1 \cdot 225}{2 (36) \cdot 39})

$A = \arccos (\frac{1296 + 1521 - 1 \cdot 225}{2 (36) \cdot 39})$

Step 4.1.4

Multiply

- 1

$- 1$ by

225

$225$ .

A = arccos (\frac{1296 + 1521 - 225}{2 (36) \cdot 39})

$A = \arccos (\frac{1296 + 1521 - 225}{2 (36) \cdot 39})$

Step 4.1.5

Add

1296

$1296$ and

1521

$1521$ .

A = arccos (\frac{2817 - 225}{2 (36) \cdot 39})

$A = \arccos (\frac{2817 - 225}{2 (36) \cdot 39})$

Step 4.1.6

Subtract

225

$225$ from

2817

$2817$ .

A = arccos (\frac{2592}{2 (36) \cdot 39})

$A = \arccos (\frac{2592}{2 (36) \cdot 39})$

A = arccos (\frac{2592}{2 (36) \cdot 39})

$A = \arccos (\frac{2592}{2 (36) \cdot 39})$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply

2

$2$ by

36

$36$ .

A = arccos (\frac{2592}{72 \cdot 39})

$A = \arccos (\frac{2592}{72 \cdot 39})$

Step 4.2.2

Multiply

72

$72$ by

39

$39$ .

A = arccos (\frac{2592}{2808})

$A = \arccos (\frac{2592}{2808})$

A = arccos (\frac{2592}{2808})

$A = \arccos (\frac{2592}{2808})$

Step 4.3

Cancel the common factor of

2592

$2592$ and

2808

$2808$ .

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Step 4.3.1

Factor

216

$216$ out of

2592

$2592$ .

A = arccos (\frac{216 (12)}{2808})

$A = \arccos (\frac{216 (12)}{2808})$

Step 4.3.2

Cancel the common factors.

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Step 4.3.2.1

Factor

216

$216$ out of

2808

$2808$ .

A = arccos (\frac{216 \cdot 12}{216 \cdot 13})

$A = \arccos (\frac{216 \cdot 12}{216 \cdot 13})$

Step 4.3.2.2

Cancel the common factor.

$A = \arccos (\frac{216 \cdot 12}{216 \cdot 13})$

Step 4.3.2.3

Rewrite the expression.

$A = \arccos (\frac{12}{13})$

Step 4.4

Evaluate

$\arccos (\frac{12}{13})$ .

$A = 22.61986494$

Step 5

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

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

Step 6

Solve the equation.

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

Step 7

Substitute the known values into the equation.

$B = \arccos (\frac{{(15)}^{2} + {(39)}^{2} - {(36)}^{2}}{2 (15) (39)})$

Step 8

Simplify the results.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Raise

$15$ to the power of

$2$ .

$B = \arccos (\frac{225 + 39^{2} - 36^{2}}{2 (15) \cdot 39})$

Step 8.1.2

Raise

$39$ to the power of

$2$ .

$B = \arccos (\frac{225 + 1521 - 36^{2}}{2 (15) \cdot 39})$

Step 8.1.3

Raise

$36$ to the power of

$2$ .

$B = \arccos (\frac{225 + 1521 - 1 \cdot 1296}{2 (15) \cdot 39})$

Step 8.1.4

Multiply

$- 1$ by

$1296$ .

$B = \arccos (\frac{225 + 1521 - 1296}{2 (15) \cdot 39})$

Step 8.1.5

Add

$225$ and

$1521$ .

$B = \arccos (\frac{1746 - 1296}{2 (15) \cdot 39})$

Step 8.1.6

Subtract

$1296$ from

$1746$ .

$B = \arccos (\frac{450}{2 (15) \cdot 39})$

Step 8.2

Simplify the denominator.

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Step 8.2.1

Multiply

$2$ by

$15$ .

$B = \arccos (\frac{450}{30 \cdot 39})$

Step 8.2.2

Multiply

$30$ by

$39$ .

$B = \arccos (\frac{450}{1170})$

Step 8.3

Cancel the common factor of

$450$ and

$1170$ .

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Step 8.3.1

Factor

$90$ out of

$450$ .

$B = \arccos (\frac{90 (5)}{1170})$

Step 8.3.2

Cancel the common factors.

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Step 8.3.2.1

Factor

$90$ out of

$1170$ .

$B = \arccos (\frac{90 \cdot 5}{90 \cdot 13})$

Step 8.3.2.2

Cancel the common factor.

$B = \arccos (\frac{90 \cdot 5}{90 \cdot 13})$

Step 8.3.2.3

Rewrite the expression.

$B = \arccos (\frac{5}{13})$

Step 8.4

Evaluate

$\arccos (\frac{5}{13})$ .

$B = 67.38013505$

Step 9

The sum of all the angles in a triangle is

$180$ degrees.

$22.61986494 + C + 67.38013505 = 180$

Step 10

Solve the equation for

$C$ .

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Step 10.1

Add

$22.61986494$ and

$67.38013505$ .

$C + 90 = 180$

Step 10.2

Move all terms not containing

$C$ to the right side of the equation.

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Step 10.2.1

Subtract

$90$ from both sides of the equation.

$C = 180 - 90$

Step 10.2.2

Subtract

$90$ from

$180$ .

$C = 90$

Step 11

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

$A = 22.61986494$

$B = 67.38013505$

$C = 90$

$a = 15$

$b = 36$

$c = 39$