Trigonometry Examples

b = 60.79

$b = 60.79$ ,

c = 60.79

$c = 60.79$ ,

C = 70.89

$C = 70.89$

Step 1

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 2

Substitute the known values into the law of sines to find

B

$B$ .

\frac{sin (B)}{60.79} = \frac{sin (70.89)}{60.79}

$\frac{\sin (B)}{60.79} = \frac{\sin (70.89)}{60.79}$

Step 3

Solve the equation for

B

$B$ .

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Step 3.1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

sin (B) = sin (70.89)

$\sin (B) = \sin (70.89)$

Step 3.2

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

B = 70.89

$B = 70.89$

Step 3.3

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 - 70.89

$B = 180 - 70.89$

Step 3.4

Subtract

70.89

$70.89$ from

180

$180$ .

B = 109.11

$B = 109.11$

Step 3.5

Find the period of

sin (B)

$\sin (B)$ .

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Step 3.5.1

The period of the function can be calculated using

\frac{360}{| b |}

$\frac{360}{| b |}$ .

\frac{360}{| b |}

$\frac{360}{| b |}$

Step 3.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{360}{| 1 |}

$\frac{360}{| 1 |}$

Step 3.5.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{360}{1}

$\frac{360}{1}$

Step 3.5.4

Divide

360

$360$ by

1

$1$ .

360

$360$

360

$360$

Step 3.6

The period of the

sin (B)

$\sin (B)$ function is

360

$360$ so values will repeat every

360

$360$ degrees in both directions.

B = 70.89 + 360 n, 109.11 + 360 n

$B = 70.89 + 360 n, 109.11 + 360 n$ , for any integer

n

$n$

Step 3.7

The triangle is invalid.

Invalid triangle

Step 4

\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 5

Substitute the known values into the law of sines to find

B

$B$ .

\frac{sin (B)}{60.79} = \frac{sin (70.89)}{60.79}

$\frac{\sin (B)}{60.79} = \frac{\sin (70.89)}{60.79}$

Step 6

Solve the equation for

B

$B$ .

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Step 6.1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

sin (B) = sin (70.89)

$\sin (B) = \sin (70.89)$

Step 6.2

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

B = 70.89

$B = 70.89$

Step 6.3

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 - 70.89

$B = 180 - 70.89$

Step 6.4

Subtract

70.89

$70.89$ from

180

$180$ .

B = 109.11

$B = 109.11$

Step 6.5

Find the period of

sin (B)

$\sin (B)$ .

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Step 6.5.1

The period of the function can be calculated using

\frac{360}{| b |}

$\frac{360}{| b |}$ .

\frac{360}{| b |}

$\frac{360}{| b |}$

Step 6.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{360}{| 1 |}

$\frac{360}{| 1 |}$

Step 6.5.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{360}{1}

$\frac{360}{1}$

Step 6.5.4

Divide

360

$360$ by

1

$1$ .

360

$360$

360

$360$

Step 6.6

The period of the

sin (B)

$\sin (B)$ function is

360

$360$ so values will repeat every

360

$360$ degrees in both directions.

B = 70.89 + 360 n, 109.11 + 360 n

$B = 70.89 + 360 n, 109.11 + 360 n$ , for any integer

n

$n$

Step 6.7

The triangle is invalid.

Invalid triangle

Step 7

\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 8

Substitute the known values into the law of sines to find

B

$B$ .

\frac{sin (B)}{60.79} = \frac{sin (70.89)}{60.79}

$\frac{\sin (B)}{60.79} = \frac{\sin (70.89)}{60.79}$

Step 9

Solve the equation for

B

$B$ .

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Step 9.1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

sin (B) = sin (70.89)

$\sin (B) = \sin (70.89)$

Step 9.2

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

B = 70.89

$B = 70.89$

Step 9.3

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 - 70.89

$B = 180 - 70.89$

Step 9.4

Subtract

70.89

$70.89$ from

180

$180$ .

B = 109.11

$B = 109.11$

Step 9.5

Find the period of

sin (B)

$\sin (B)$ .

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Step 9.5.1

The period of the function can be calculated using

\frac{360}{| b |}

$\frac{360}{| b |}$ .

\frac{360}{| b |}

$\frac{360}{| b |}$

Step 9.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{360}{| 1 |}

$\frac{360}{| 1 |}$

Step 9.5.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{360}{1}

$\frac{360}{1}$

Step 9.5.4

Divide

360

$360$ by

1

$1$ .

360

$360$

360

$360$

Step 9.6

The period of the

sin (B)

$\sin (B)$ function is

360

$360$ so values will repeat every

360

$360$ degrees in both directions.

B = 70.89 + 360 n, 109.11 + 360 n

$B = 70.89 + 360 n, 109.11 + 360 n$ , for any integer

n

$n$

Step 9.7

The triangle is invalid.

Invalid triangle

Step 10

\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 11

Substitute the known values into the law of sines to find

B

$B$ .

\frac{sin (B)}{60.79} = \frac{sin (70.89)}{60.79}

$\frac{\sin (B)}{60.79} = \frac{\sin (70.89)}{60.79}$

Step 12

Solve the equation for

B

$B$ .

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Step 12.1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

sin (B) = sin (70.89)

$\sin (B) = \sin (70.89)$

Step 12.2

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

B = 70.89

$B = 70.89$

Step 12.3

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 - 70.89

$B = 180 - 70.89$

Step 12.4

Subtract

70.89

$70.89$ from

180

$180$ .

B = 109.11

$B = 109.11$

Step 12.5

Find the period of

sin (B)

$\sin (B)$ .

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Step 12.5.1

The period of the function can be calculated using

\frac{360}{| b |}

$\frac{360}{| b |}$ .

\frac{360}{| b |}

$\frac{360}{| b |}$

Step 12.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{360}{| 1 |}

$\frac{360}{| 1 |}$

Step 12.5.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{360}{1}

$\frac{360}{1}$

Step 12.5.4

Divide

360

$360$ by

1

$1$ .

360

$360$

360

$360$

Step 12.6

The period of the

sin (B)

$\sin (B)$ function is

360

$360$ so values will repeat every

360

$360$ degrees in both directions.

B = 70.89 + 360 n, 109.11 + 360 n

$B = 70.89 + 360 n, 109.11 + 360 n$ , for any integer

n

$n$

Step 12.7

The triangle is invalid.

Invalid 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)}{60.79} = \frac{sin (70.89)}{60.79}

$\frac{\sin (B)}{60.79} = \frac{\sin (70.89)}{60.79}$

Step 15

Solve the equation for

B

$B$ .

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Step 15.1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

sin (B) = sin (70.89)

$\sin (B) = \sin (70.89)$

Step 15.2

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

B = 70.89

$B = 70.89$

Step 15.3

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 - 70.89

$B = 180 - 70.89$

Step 15.4

Subtract

70.89

$70.89$ from

180

$180$ .

B = 109.11

$B = 109.11$

Step 15.5

Find the period of

sin (B)

$\sin (B)$ .

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Step 15.5.1

The period of the function can be calculated using

\frac{360}{| b |}

$\frac{360}{| b |}$ .

\frac{360}{| b |}

$\frac{360}{| b |}$

Step 15.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{360}{| 1 |}

$\frac{360}{| 1 |}$

Step 15.5.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{360}{1}

$\frac{360}{1}$

Step 15.5.4

Divide

360

$360$ by

1

$1$ .

360

$360$

360

$360$

Step 15.6

The period of the

sin (B)

$\sin (B)$ function is

360

$360$ so values will repeat every

360

$360$ degrees in both directions.

B = 70.89 + 360 n, 109.11 + 360 n

$B = 70.89 + 360 n, 109.11 + 360 n$ , for any integer

n

$n$

Step 15.7

The triangle is invalid.

Invalid triangle

Step 16

\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 17

Substitute the known values into the law of sines to find

B

$B$ .

\frac{sin (B)}{60.79} = \frac{sin (70.89)}{60.79}

$\frac{\sin (B)}{60.79} = \frac{\sin (70.89)}{60.79}$

Step 18

Solve the equation for

B

$B$ .

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Step 18.1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

sin (B) = sin (70.89)

$\sin (B) = \sin (70.89)$

Step 18.2

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

B = 70.89

$B = 70.89$

Step 18.3

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 - 70.89

$B = 180 - 70.89$

Step 18.4

Subtract

70.89

$70.89$ from

180

$180$ .

B = 109.11

$B = 109.11$

Step 18.5

Find the period of

sin (B)

$\sin (B)$ .

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Step 18.5.1

The period of the function can be calculated using

\frac{360}{| b |}

$\frac{360}{| b |}$ .

\frac{360}{| b |}

$\frac{360}{| b |}$

Step 18.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{360}{| 1 |}

$\frac{360}{| 1 |}$

Step 18.5.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{360}{1}

$\frac{360}{1}$

Step 18.5.4

Divide

360

$360$ by

1

$1$ .

360

$360$

360

$360$

Step 18.6

The period of the

sin (B)

$\sin (B)$ function is

360

$360$ so values will repeat every

360

$360$ degrees in both directions.

B = 70.89 + 360 n, 109.11 + 360 n

$B = 70.89 + 360 n, 109.11 + 360 n$ , for any integer

n

$n$

Step 18.7

The triangle is invalid.

Invalid triangle

Step 19

There are not enough parameters given to solve the triangle.

Unknown triangle