Trigonometry Examples

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\begin{matrix} Side & Angle \begin{matrix} b = c = & 6 a = & 4 \end{matrix} & \begin{matrix} A = B = C = \end{matrix} \end{matrix}

$\begin{matrix} Side & Angle \\ \begin{matrix} b = \\ c = & 6 \\ a = & 4 \end{matrix} & \begin{matrix} A = \\ B = \\ C = \end{matrix} \end{matrix}$

Step 1

Assume that angle

C = 90

$C = 90$ .

C = 90

$C = 90$

Step 2

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

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Step 2.1

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 2.2

Solve the equation for

b

$b$ .

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

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

Step 2.3

Substitute the actual values into the equation.

b = \sqrt{{(6)}^{2} - {(4)}^{2}}

$b = \sqrt{{(6)}^{2} - {(4)}^{2}}$

Step 2.4

Raise

6

$6$ to the power of

2

$2$ .

b = \sqrt{36 - {(4)}^{2}}

$b = \sqrt{36 - {(4)}^{2}}$

Step 2.5

Raise

4

$4$ to the power of

2

$2$ .

b = \sqrt{36 - 1 \cdot 16}

$b = \sqrt{36 - 1 \cdot 16}$

Step 2.6

Multiply

- 1

$- 1$ by

16

$16$ .

b = \sqrt{36 - 16}

$b = \sqrt{36 - 16}$

Step 2.7

Subtract

16

$16$ from

36

$36$ .

b = \sqrt{20}

$b = \sqrt{20}$

Step 2.8

Rewrite

20

$20$ as

2^{2} \cdot 5

$2^{2} \cdot 5$ .

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Step 2.8.1

Factor

4

$4$ out of

20

$20$ .

b = \sqrt{4 (5)}

$b = \sqrt{4 (5)}$

Step 2.8.2

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

b = \sqrt{2^{2} \cdot 5}

$b = \sqrt{2^{2} \cdot 5}$

b = \sqrt{2^{2} \cdot 5}

$b = \sqrt{2^{2} \cdot 5}$

Step 2.9

Pull terms out from under the radical.

b = 2 \sqrt{5}

$b = 2 \sqrt{5}$

b = 2 \sqrt{5}

$b = 2 \sqrt{5}$

Step 3

Find

B

$B$ .

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Step 3.1

The angle

B

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

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

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

Step 3.2

Substitute in the values of the opposite side to angle

B

$B$ and hypotenuse

6

$6$ of the triangle.

B = arcsin (\frac{2 \sqrt{5}}{6})

$B = \arcsin (\frac{2 \sqrt{5}}{6})$

Step 3.3

Cancel the common factor of

2

$2$ and

6

$6$ .

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Step 3.3.1

Factor

2

$2$ out of

2 \sqrt{5}

$2 \sqrt{5}$ .

B = arcsin ⎛ ⎜ ⎝ \frac{2 (\sqrt{5})}{6} ⎞ ⎟ ⎠

$B = \arcsin (\frac{2 (\sqrt{5})}{6})$

Step 3.3.2

Cancel the common factors.

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Step 3.3.2.1

Factor

2

$2$ out of

6

$6$ .

B = arcsin (\frac{2 \sqrt{5}}{2 \cdot 3})

$B = \arcsin (\frac{2 \sqrt{5}}{2 \cdot 3})$

Step 3.3.2.2

Cancel the common factor.

$B = \arcsin (\frac{2 \sqrt{5}}{2 \cdot 3})$

Step 3.3.2.3

Rewrite the expression.

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

Step 3.4

Evaluate

$\arcsin (\frac{\sqrt{5}}{3})$ .

$B = 48.1896851$

Step 4

Find the last angle of the triangle.

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Step 4.1

The sum of all the angles in a triangle is

$180$ degrees.

$A + 90 + 48.1896851 = 180$

Step 4.2

Solve the equation for

$A$ .

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Step 4.2.1

Add

$90$ and

$48.1896851$ .

$A + 138.1896851 = 180$

Step 4.2.2

Move all terms not containing

$A$ to the right side of the equation.

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Step 4.2.2.1

Subtract

$138.1896851$ from both sides of the equation.

$A = 180 - 138.1896851$

Step 4.2.2.2

Subtract

$138.1896851$ from

$180$ .

$A = 41.81031489$

Step 5

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

$A = 41.81031489$

$B = 48.1896851$

$C = 90$

$a = 4$

$b = 2 \sqrt{5}$

$c = 6$