Trigonometry Examples

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\begin{matrix} Side & Angle \begin{matrix} b = c = & 15 a = & 9 \end{matrix} & \begin{matrix} A = B = C = & 90 \end{matrix} \end{matrix}

$\begin{matrix} Side & Angle \\ \begin{matrix} b = \\ c = & 15 \\ a = & 9 \end{matrix} & \begin{matrix} A = \\ B = \\ C = & 90 \end{matrix} \end{matrix}$

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

b

$b$ .

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

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

Step 1.3

Substitute the actual values into the equation.

b = \sqrt{{(15)}^{2} - {(9)}^{2}}

$b = \sqrt{{(15)}^{2} - {(9)}^{2}}$

Step 1.4

Raise

15

$15$ to the power of

2

$2$ .

b = \sqrt{225 - {(9)}^{2}}

$b = \sqrt{225 - {(9)}^{2}}$

Step 1.5

Raise

9

$9$ to the power of

2

$2$ .

b = \sqrt{225 - 1 \cdot 81}

$b = \sqrt{225 - 1 \cdot 81}$

Step 1.6

Multiply

- 1

$- 1$ by

81

$81$ .

b = \sqrt{225 - 81}

$b = \sqrt{225 - 81}$

Step 1.7

Subtract

81

$81$ from

225

$225$ .

b = \sqrt{144}

$b = \sqrt{144}$

Step 1.8

Rewrite

144

$144$ as

12^{2}

$12^{2}$ .

b = \sqrt{12^{2}}

$b = \sqrt{12^{2}}$

Step 1.9

Pull terms out from under the radical, assuming positive real numbers.

b = 12

$b = 12$

b = 12

$b = 12$

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

15

$15$ of the triangle.

B = arcsin (\frac{12}{15})

$B = \arcsin (\frac{12}{15})$

Step 2.3

Cancel the common factor of

12

$12$ and

15

$15$ .

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Step 2.3.1

Factor

3

$3$ out of

12

$12$ .

B = arcsin (\frac{3 (4)}{15})

$B = \arcsin (\frac{3 (4)}{15})$

Step 2.3.2

Cancel the common factors.

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Step 2.3.2.1

Factor

3

$3$ out of

15

$15$ .

B = arcsin (\frac{3 \cdot 4}{3 \cdot 5})

$B = \arcsin (\frac{3 \cdot 4}{3 \cdot 5})$

Step 2.3.2.2

Cancel the common factor.

$B = \arcsin (\frac{3 \cdot 4}{3 \cdot 5})$

Step 2.3.2.3

Rewrite the expression.

$B = \arcsin (\frac{4}{5})$

Step 2.4

Evaluate

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

$B = 53.13010235$

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.

$A + 90 + 53.13010235 = 180$

Step 3.2

Solve the equation for

$A$ .

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Step 3.2.1

Add

$90$ and

$53.13010235$ .

$A + 143.13010235 = 180$

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

$143.13010235$ from both sides of the equation.

$A = 180 - 143.13010235$

Step 3.2.2.2

Subtract

$143.13010235$ from

$180$ .

$A = 36.86989764$

Step 4

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

$A = 36.86989764$

$B = 53.13010235$

$C = 90$

$a = 9$

$b = 12$

$c = 15$