Trigonometry Examples

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\begin{matrix} Side & Angle \begin{matrix} b = & 16 c = & 20 a = & 12 \end{matrix} & \begin{matrix} A = B = C = \end{matrix} \end{matrix}

$\begin{matrix} Side & Angle \\ \begin{matrix} b = & 16 \\ c = & 20 \\ a = & 12 \end{matrix} & \begin{matrix} A = \\ B = \\ C = \end{matrix} \end{matrix}$

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{{(16)}^{2} + {(20)}^{2} - {(12)}^{2}}{2 (16) (20)})

$A = \arccos (\frac{{(16)}^{2} + {(20)}^{2} - {(12)}^{2}}{2 (16) (20)})$

Step 4

Simplify the results.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise

16

$16$ to the power of

2

$2$ .

A = arccos (\frac{256 + 20^{2} - 12^{2}}{2 (16) \cdot 20})

$A = \arccos (\frac{256 + 20^{2} - 12^{2}}{2 (16) \cdot 20})$

Step 4.1.2

Raise

20

$20$ to the power of

2

$2$ .

A = arccos (\frac{256 + 400 - 12^{2}}{2 (16) \cdot 20})

$A = \arccos (\frac{256 + 400 - 12^{2}}{2 (16) \cdot 20})$

Step 4.1.3

Raise

12

$12$ to the power of

2

$2$ .

A = arccos (\frac{256 + 400 - 1 \cdot 144}{2 (16) \cdot 20})

$A = \arccos (\frac{256 + 400 - 1 \cdot 144}{2 (16) \cdot 20})$

Step 4.1.4

Multiply

- 1

$- 1$ by

144

$144$ .

A = arccos (\frac{256 + 400 - 144}{2 (16) \cdot 20})

$A = \arccos (\frac{256 + 400 - 144}{2 (16) \cdot 20})$

Step 4.1.5

Add

256

$256$ and

400

$400$ .

A = arccos (\frac{656 - 144}{2 (16) \cdot 20})

$A = \arccos (\frac{656 - 144}{2 (16) \cdot 20})$

Step 4.1.6

Subtract

144

$144$ from

656

$656$ .

A = arccos (\frac{512}{2 (16) \cdot 20})

$A = \arccos (\frac{512}{2 (16) \cdot 20})$

A = arccos (\frac{512}{2 (16) \cdot 20})

$A = \arccos (\frac{512}{2 (16) \cdot 20})$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply

2

$2$ by

16

$16$ .

A = arccos (\frac{512}{32 \cdot 20})

$A = \arccos (\frac{512}{32 \cdot 20})$

Step 4.2.2

Multiply

32

$32$ by

20

$20$ .

A = arccos (\frac{512}{640})

$A = \arccos (\frac{512}{640})$

A = arccos (\frac{512}{640})

$A = \arccos (\frac{512}{640})$

Step 4.3

Cancel the common factor of

512

$512$ and

640

$640$ .

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Step 4.3.1

Factor

128

$128$ out of

512

$512$ .

A = arccos (\frac{128 (4)}{640})

$A = \arccos (\frac{128 (4)}{640})$

Step 4.3.2

Cancel the common factors.

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Step 4.3.2.1

Factor

128

$128$ out of

640

$640$ .

A = arccos (\frac{128 \cdot 4}{128 \cdot 5})

$A = \arccos (\frac{128 \cdot 4}{128 \cdot 5})$

Step 4.3.2.2

Cancel the common factor.

$A = \arccos (\frac{128 \cdot 4}{128 \cdot 5})$

Step 4.3.2.3

Rewrite the expression.

$A = \arccos (\frac{4}{5})$

Step 4.4

Evaluate

$\arccos (\frac{4}{5})$ .

$A = 36.86989764$

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.

$B = \arccos (\frac{{(12)}^{2} + {(20)}^{2} - {(16)}^{2}}{2 (12) (20)})$

Step 8

Simplify the results.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Raise

$12$ to the power of

$2$ .

$B = \arccos (\frac{144 + 20^{2} - 16^{2}}{2 (12) \cdot 20})$

Step 8.1.2

Raise

$20$ to the power of

$2$ .

$B = \arccos (\frac{144 + 400 - 16^{2}}{2 (12) \cdot 20})$

Step 8.1.3

Raise

$16$ to the power of

$2$ .

$B = \arccos (\frac{144 + 400 - 1 \cdot 256}{2 (12) \cdot 20})$

Step 8.1.4

Multiply

$- 1$ by

$256$ .

$B = \arccos (\frac{144 + 400 - 256}{2 (12) \cdot 20})$

Step 8.1.5

Add

$144$ and

$400$ .

$B = \arccos (\frac{544 - 256}{2 (12) \cdot 20})$

Step 8.1.6

Subtract

$256$ from

$544$ .

$B = \arccos (\frac{288}{2 (12) \cdot 20})$

Step 8.2

Simplify the denominator.

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Step 8.2.1

Multiply

$2$ by

$12$ .

$B = \arccos (\frac{288}{24 \cdot 20})$

Step 8.2.2

Multiply

$24$ by

$20$ .

$B = \arccos (\frac{288}{480})$

Step 8.3

Cancel the common factor of

$288$ and

$480$ .

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Step 8.3.1

Factor

$96$ out of

$288$ .

$B = \arccos (\frac{96 (3)}{480})$

Step 8.3.2

Cancel the common factors.

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Step 8.3.2.1

Factor

$96$ out of

$480$ .

$B = \arccos (\frac{96 \cdot 3}{96 \cdot 5})$

Step 8.3.2.2

Cancel the common factor.

$B = \arccos (\frac{96 \cdot 3}{96 \cdot 5})$

Step 8.3.2.3

Rewrite the expression.

$B = \arccos (\frac{3}{5})$

Step 8.4

Evaluate

$\arccos (\frac{3}{5})$ .

$B = 53.13010235$

Step 9

The sum of all the angles in a triangle is

$180$ degrees.

$36.86989764 + C + 53.13010235 = 180$

Step 10

Solve the equation for

$C$ .

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Step 10.1

Add

$36.86989764$ and

$53.13010235$ .

$C + 90 = 180$

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.

$A = 36.86989764$

$B = 53.13010235$

$C = 90$

$a = 12$

$b = 16$

$c = 20$