Trigonometry Examples

$a = 40$ , $b = 27$ , $c = 16$

Step 1

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

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

Step 2

Solve the equation.

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

Step 3

Substitute the known values into the equation.

$A = \arccos (\frac{{(27)}^{2} + {(16)}^{2} - {(40)}^{2}}{2 (27) (16)})$

Step 4

Simplify the results.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $27$ to the power of $2$ .

$A = \arccos (\frac{729 + 16^{2} - 40^{2}}{2 (27) \cdot 16})$

Step 4.1.2

Raise $16$ to the power of $2$ .

$A = \arccos (\frac{729 + 256 - 40^{2}}{2 (27) \cdot 16})$

Step 4.1.3

Raise $40$ to the power of $2$ .

$A = \arccos (\frac{729 + 256 - 1 \cdot 1600}{2 (27) \cdot 16})$

Step 4.1.4

Multiply $- 1$ by $1600$ .

$A = \arccos (\frac{729 + 256 - 1600}{2 (27) \cdot 16})$

Step 4.1.5

Add $729$ and $256$ .

$A = \arccos (\frac{985 - 1600}{2 (27) \cdot 16})$

Step 4.1.6

Subtract $1600$ from $985$ .

$A = \arccos (\frac{- 615}{2 (27) \cdot 16})$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $2$ by $27$ .

$A = \arccos (\frac{- 615}{54 \cdot 16})$

Step 4.2.2

Multiply $54$ by $16$ .

$A = \arccos (\frac{- 615}{864})$

Step 4.3

Reduce the expression by cancelling the common factors.

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Step 4.3.1

Cancel the common factor of $- 615$ and $864$ .

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Step 4.3.1.1

Factor $3$ out of $- 615$ .

$A = \arccos (\frac{3 (- 205)}{864})$

Step 4.3.1.2

Cancel the common factors.

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Step 4.3.1.2.1

Factor $3$ out of $864$ .

$A = \arccos (\frac{3 \cdot - 205}{3 \cdot 288})$

Step 4.3.1.2.2

Cancel the common factor.

$A = \arccos (\frac{3 \cdot - 205}{3 \cdot 288})$

Step 4.3.1.2.3

Rewrite the expression.

$A = \arccos (\frac{- 205}{288})$

Step 4.3.2

Move the negative in front of the fraction.

$A = \arccos (- \frac{205}{288})$

Step 4.4

Evaluate $\arccos (- \frac{205}{288})$ .

$A = 135.38201081$

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.

$B = \arccos (\frac{{(40)}^{2} + {(16)}^{2} - {(27)}^{2}}{2 (40) (16)})$

Step 8

Simplify the results.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Raise $40$ to the power of $2$ .

$B = \arccos (\frac{1600 + 16^{2} - 27^{2}}{2 (40) \cdot 16})$

Step 8.1.2

Raise $16$ to the power of $2$ .

$B = \arccos (\frac{1600 + 256 - 27^{2}}{2 (40) \cdot 16})$

Step 8.1.3

Raise $27$ to the power of $2$ .

$B = \arccos (\frac{1600 + 256 - 1 \cdot 729}{2 (40) \cdot 16})$

Step 8.1.4

Multiply $- 1$ by $729$ .

$B = \arccos (\frac{1600 + 256 - 729}{2 (40) \cdot 16})$

Step 8.1.5

Add $1600$ and $256$ .

$B = \arccos (\frac{1856 - 729}{2 (40) \cdot 16})$

Step 8.1.6

Subtract $729$ from $1856$ .

$B = \arccos (\frac{1127}{2 (40) \cdot 16})$

Step 8.2

Simplify the denominator.

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Step 8.2.1

Multiply $2$ by $40$ .

$B = \arccos (\frac{1127}{80 \cdot 16})$

Step 8.2.2

Multiply $80$ by $16$ .

$B = \arccos (\frac{1127}{1280})$

Step 8.3

Evaluate $\arccos (\frac{1127}{1280})$ .

$B = 28.30103975$

Step 9

The sum of all the angles in a triangle is $180$ degrees.

$135.38201081 + C + 28.30103975 = 180$

Step 10

Solve the equation for $C$ .

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Step 10.1

Add $135.38201081$ and $28.30103975$ .

$C + 163.68305057 = 180$

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 $163.68305057$ from both sides of the equation.

$C = 180 - 163.68305057$

Step 10.2.2

Subtract $163.68305057$ from $180$ .

$C = 16.31694942$

Step 11

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

$A = 135.38201081$

$B = 28.30103975$

$C = 16.31694942$

$a = 40$

$b = 27$

$c = 16$