Trigonometry Examples

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\begin{matrix} Side & Angle \begin{matrix} b = & 5 c = & 10 a = \end{matrix} & \begin{matrix} A = B = C = & 90 \end{matrix} \end{matrix}

$\begin{matrix} Side & Angle \\ \begin{matrix} b = & 5 \\ c = & 10 \\ a = \end{matrix} & \begin{matrix} A = \\ B = \\ C = & 90 \end{matrix} \end{matrix}$

Step 1

Find the last side of the triangle using the Pythagorean theorem.

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Step 1.1

Use the Pythagorean theorem to find the unknown side. In any right triangle, the area of the square whose side is the hypotenuse (the side of a right triangle opposite the right angle) is equal to the sum of areas of the squares whose sides are the two legs (the two sides other than the hypotenuse).

a^{2} + b^{2} = c^{2}

$a^{2} + b^{2} = c^{2}$

Step 1.2

Solve the equation for

a

$a$ .

a = \sqrt{c^{2} - b^{2}}

$a = \sqrt{c^{2} - b^{2}}$

Step 1.3

Substitute the actual values into the equation.

a = \sqrt{{(10)}^{2} - {(5)}^{2}}

$a = \sqrt{{(10)}^{2} - {(5)}^{2}}$

Step 1.4

Raise

10

$10$ to the power of

2

$2$ .

a = \sqrt{100 - {(5)}^{2}}

$a = \sqrt{100 - {(5)}^{2}}$

Step 1.5

Raise

5

$5$ to the power of

2

$2$ .

a = \sqrt{100 - 1 \cdot 25}

$a = \sqrt{100 - 1 \cdot 25}$

Step 1.6

Multiply

- 1

$- 1$ by

25

$25$ .

a = \sqrt{100 - 25}

$a = \sqrt{100 - 25}$

Step 1.7

Subtract

25

$25$ from

100

$100$ .

a = \sqrt{75}

$a = \sqrt{75}$

Step 1.8

Rewrite

75

$75$ as

5^{2} \cdot 3

$5^{2} \cdot 3$ .

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Step 1.8.1

Factor

25

$25$ out of

75

$75$ .

a = \sqrt{25 (3)}

$a = \sqrt{25 (3)}$

Step 1.8.2

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

a = \sqrt{5^{2} \cdot 3}

$a = \sqrt{5^{2} \cdot 3}$

a = \sqrt{5^{2} \cdot 3}

$a = \sqrt{5^{2} \cdot 3}$

Step 1.9

Pull terms out from under the radical.

a = 5 \sqrt{3}

$a = 5 \sqrt{3}$

a = 5 \sqrt{3}

$a = 5 \sqrt{3}$

Step 2

Find

B

$B$ .

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Step 2.1

The angle

B

$B$ can be found using the inverse sine function.

B = arcsin (\frac{opp}{hyp})

$B = \arcsin (\frac{opp}{hyp})$

Step 2.2

Substitute in the values of the opposite side to angle

B

$B$ and hypotenuse

10

$10$ of the triangle.

B = arcsin (\frac{5}{10})

$B = \arcsin (\frac{5}{10})$

Step 2.3

Cancel the common factor of

5

$5$ and

10

$10$ .

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Step 2.3.1

Factor

5

$5$ out of

5

$5$ .

B = arcsin (\frac{5 (1)}{10})

$B = \arcsin (\frac{5 (1)}{10})$

Step 2.3.2

Cancel the common factors.

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Step 2.3.2.1

Factor

5

$5$ out of

10

$10$ .

B = arcsin (\frac{5 \cdot 1}{5 \cdot 2})

$B = \arcsin (\frac{5 \cdot 1}{5 \cdot 2})$

Step 2.3.2.2

Cancel the common factor.

$B = \arcsin (\frac{5 \cdot 1}{5 \cdot 2})$

Step 2.3.2.3

Rewrite the expression.

$B = \arcsin (\frac{1}{2})$

Step 2.4

The exact value of

$\arcsin (\frac{1}{2})$ is

$30$ .

$B = 30$

Step 3

Find the last angle of the triangle.

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Step 3.1

The sum of all the angles in a triangle is

$180$ degrees.

$A + 90 + 30 = 180$

Step 3.2

Solve the equation for

$A$ .

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Step 3.2.1

Add

$90$ and

$30$ .

$A + 120 = 180$

Step 3.2.2

Move all terms not containing

$A$ to the right side of the equation.

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Step 3.2.2.1

Subtract

$120$ from both sides of the equation.

$A = 180 - 120$

Step 3.2.2.2

Subtract

$120$ from

$180$ .

$A = 60$

Step 4

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

$A = 60$

$B = 30$

$C = 90$

$a = 5 \sqrt{3}$

$b = 5$

$c = 10$