Trigonometry Examples

b = 1

$b = 1$ ,

c = 2

$c = 2$ ,

A = 150

$A = 150$

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

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

Step 3

Substitute the known values into the equation.

a = \sqrt{{(1)}^{2} + {(2)}^{2} - 2 \cdot 1 \cdot 2 cos (150)}

$a = \sqrt{{(1)}^{2} + {(2)}^{2} - 2 \cdot 1 \cdot 2 \cos (150)}$

Step 4

Simplify the results.

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Step 4.1

One to any power is one.

a = \sqrt{1 + {(2)}^{2} - 2 \cdot 1 \cdot (2 cos (150))}

$a = \sqrt{1 + {(2)}^{2} - 2 \cdot 1 \cdot (2 \cos (150))}$

Step 4.2

Raise

2

$2$ to the power of

2

$2$ .

a = \sqrt{1 + 4 - 2 \cdot 1 \cdot (2 cos (150))}

$a = \sqrt{1 + 4 - 2 \cdot 1 \cdot (2 \cos (150))}$

Step 4.3

Multiply

- 2 \cdot 1 \cdot 2

$- 2 \cdot 1 \cdot 2$ .

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Step 4.3.1

Multiply

- 2

$- 2$ by

1

$1$ .

a = \sqrt{1 + 4 - 2 \cdot (2 cos (150))}

$a = \sqrt{1 + 4 - 2 \cdot (2 \cos (150))}$

Step 4.3.2

Multiply

- 2

$- 2$ by

2

$2$ .

a = \sqrt{1 + 4 - 4 cos (150)}

$a = \sqrt{1 + 4 - 4 \cos (150)}$

a = \sqrt{1 + 4 - 4 cos (150)}

$a = \sqrt{1 + 4 - 4 \cos (150)}$

Step 4.4

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

a = \sqrt{1 + 4 - 4 (- cos (30))}

$a = \sqrt{1 + 4 - 4 (- \cos (30))}$

Step 4.5

The exact value of

cos (30)

$\cos (30)$ is

\frac{\sqrt{3}}{2}

$\frac{\sqrt{3}}{2}$ .

a = \sqrt{1 + 4 - 4 (- \frac{\sqrt{3}}{2})}

$a = \sqrt{1 + 4 - 4 (- \frac{\sqrt{3}}{2})}$

Step 4.6

Cancel the common factor of

2

$2$ .

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Step 4.6.1

Move the leading negative in

- \frac{\sqrt{3}}{2}

$- \frac{\sqrt{3}}{2}$ into the numerator.

a = \sqrt{1 + 4 - 4 \frac{- \sqrt{3}}{2}}

$a = \sqrt{1 + 4 - 4 \frac{- \sqrt{3}}{2}}$

Step 4.6.2

Factor

2

$2$ out of

- 4

$- 4$ .

a = \sqrt{1 + 4 + 2 (- 2) (\frac{- \sqrt{3}}{2})}

$a = \sqrt{1 + 4 + 2 (- 2) (\frac{- \sqrt{3}}{2})}$

Step 4.6.3

Cancel the common factor.

a = \sqrt{1 + 4 + 2 \cdot (- 2 \frac{- \sqrt{3}}{2})}

$a = \sqrt{1 + 4 + 2 \cdot (- 2 \frac{- \sqrt{3}}{2})}$

Step 4.6.4

Rewrite the expression.

a = \sqrt{1 + 4 - 2 (- \sqrt{3})}

$a = \sqrt{1 + 4 - 2 (- \sqrt{3})}$

a = \sqrt{1 + 4 - 2 (- \sqrt{3})}

$a = \sqrt{1 + 4 - 2 (- \sqrt{3})}$

Step 4.7

Simplify the expression.

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Step 4.7.1

Multiply

- 1

$- 1$ by

- 2

$- 2$ .

a = \sqrt{1 + 4 + 2 \sqrt{3}}

$a = \sqrt{1 + 4 + 2 \sqrt{3}}$

Step 4.7.2

Add

1

$1$ and

4

$4$ .

a = \sqrt{5 + 2 \sqrt{3}}

$a = \sqrt{5 + 2 \sqrt{3}}$

a = \sqrt{5 + 2 \sqrt{3}}

$a = \sqrt{5 + 2 \sqrt{3}}$

a = \sqrt{5 + 2 \sqrt{3}}

$a = \sqrt{5 + 2 \sqrt{3}}$

Step 5

The law of sines is based on the proportionality of sides and angles in triangles. The law states that for the angles of a non-right triangle, each angle of the triangle has the same ratio of angle measure to sine value.

\frac{sin (A)}{a} = \frac{sin (B)}{b} = \frac{sin (C)}{c}

$\frac{\sin (A)}{a} = \frac{\sin (B)}{b} = \frac{\sin (C)}{c}$

Step 6

Substitute the known values into the law of sines to find

B

$B$ .

\frac{sin (B)}{1} = \frac{sin (150)}{\sqrt{5 + 2 \sqrt{3}}}

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$

Step 7

Solve the equation for

B

$B$ .

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Step 7.1

For the two functions to be equal, the arguments of each must be equal.

B = 150

$B = 150$

Step 7.2

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from

180

$180$ to find the solution in the second quadrant.

B = 180 - 150

$B = 180 - 150$

Step 7.3

Subtract

150

$150$ from

180

$180$ .

B = 30

$B = 30$

Step 7.4

The solution to the equation

\frac{sin (B)}{1} = \frac{sin (150)}{\sqrt{5 + 2 \sqrt{3}}}

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$ .

B = 150, 30

$B = 150, 30$

Step 7.5

Exclude the solutions that do not make

\frac{sin (B)}{1} = \frac{sin (150)}{\sqrt{5 + 2 \sqrt{3}}}

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$ true.

No solution

$No solution$

No solution

$No solution$

Step 8

There are not enough parameters given to solve the triangle.

Unknown triangle

Step 9

\frac{sin (A)}{a} = \frac{sin (B)}{b} = \frac{sin (C)}{c}

$\frac{\sin (A)}{a} = \frac{\sin (B)}{b} = \frac{\sin (C)}{c}$

Step 10

Substitute the known values into the law of sines to find

B

$B$ .

\frac{sin (B)}{1} = \frac{sin (150)}{\sqrt{5 + 2 \sqrt{3}}}

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$

Step 11

Solve the equation for

B

$B$ .

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Step 11.1

For the two functions to be equal, the arguments of each must be equal.

B = 150

$B = 150$

Step 11.2

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from

180

$180$ to find the solution in the second quadrant.

B = 180 - 150

$B = 180 - 150$

Step 11.3

Subtract

150

$150$ from

180

$180$ .

B = 30

$B = 30$

Step 11.4

The solution to the equation

\frac{sin (B)}{1} = \frac{sin (150)}{\sqrt{5 + 2 \sqrt{3}}}

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$ .

B = 150, 30

$B = 150, 30$

Step 11.5

Exclude the solutions that do not make

\frac{sin (B)}{1} = \frac{sin (150)}{\sqrt{5 + 2 \sqrt{3}}}

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$ true.

No solution

$No solution$

No solution

$No solution$

Step 12

There are not enough parameters given to solve the triangle.

Unknown triangle

Step 13

\frac{sin (A)}{a} = \frac{sin (B)}{b} = \frac{sin (C)}{c}

$\frac{\sin (A)}{a} = \frac{\sin (B)}{b} = \frac{\sin (C)}{c}$

Step 14

Substitute the known values into the law of sines to find

B

$B$ .

\frac{sin (B)}{1} = \frac{sin (150)}{\sqrt{5 + 2 \sqrt{3}}}

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$

Step 15

Solve the equation for

B

$B$ .

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Step 15.1

For the two functions to be equal, the arguments of each must be equal.

B = 150

$B = 150$

Step 15.2

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from

180

$180$ to find the solution in the second quadrant.

B = 180 - 150

$B = 180 - 150$

Step 15.3

Subtract

150

$150$ from

180

$180$ .

B = 30

$B = 30$

Step 15.4

The solution to the equation

\frac{sin (B)}{1} = \frac{sin (150)}{\sqrt{5 + 2 \sqrt{3}}}

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$ .

B = 150, 30

$B = 150, 30$

Step 15.5

Exclude the solutions that do not make

\frac{sin (B)}{1} = \frac{sin (150)}{\sqrt{5 + 2 \sqrt{3}}}

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$ true.

No solution

$No solution$

No solution

$No solution$

Step 16

There are not enough parameters given to solve the triangle.

Unknown triangle

Step 17

\frac{sin (A)}{a} = \frac{sin (B)}{b} = \frac{sin (C)}{c}

$\frac{\sin (A)}{a} = \frac{\sin (B)}{b} = \frac{\sin (C)}{c}$

Step 18

Substitute the known values into the law of sines to find

B

$B$ .

\frac{sin (B)}{1} = \frac{sin (150)}{\sqrt{5 + 2 \sqrt{3}}}

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$

Step 19

Solve the equation for

$B$ .

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Step 19.1

For the two functions to be equal, the arguments of each must be equal.

$B = 150$

Step 19.2

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from

$180$ to find the solution in the second quadrant.

$B = 180 - 150$

Step 19.3

Subtract

$150$ from

$180$ .

$B = 30$

Step 19.4

The solution to the equation

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$ .

$B = 150, 30$

Step 19.5

Exclude the solutions that do not make

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$ true.

$No solution$

Step 20

There are not enough parameters given to solve the triangle.

Unknown triangle

Step 21

$\frac{\sin (A)}{a} = \frac{\sin (B)}{b} = \frac{\sin (C)}{c}$

Step 22

Substitute the known values into the law of sines to find

$B$ .

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$

Step 23

Solve the equation for

$B$ .

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Step 23.1

For the two functions to be equal, the arguments of each must be equal.

$B = 150$

Step 23.2

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from

$180$ to find the solution in the second quadrant.

$B = 180 - 150$

Step 23.3

Subtract

$150$ from

$180$ .

$B = 30$

Step 23.4

The solution to the equation

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$ .

$B = 150, 30$

Step 23.5

Exclude the solutions that do not make

$\frac{\sin (B)}{1} = \frac{\sin (150)}{\sqrt{5 + 2 \sqrt{3}}}$ true.

$No solution$

Step 24

There are not enough parameters given to solve the triangle.

Unknown triangle