Trigonometry Examples

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\begin{matrix} Side & Angle \begin{matrix} b = & 4 c = a = & 3 \end{matrix} & \begin{matrix} A = B = C = & 90 \end{matrix} \end{matrix}

$\begin{matrix} Side & Angle \\ \begin{matrix} b = & 4 \\ c = \\ a = & 3 \end{matrix} & \begin{matrix} A = \\ B = \\ C = & 90 \end{matrix} \end{matrix}$

Step 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

Solve the equation for

c

$c$ .

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

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

Step 3

Substitute the actual values into the equation.

c = \sqrt{{(4)}^{2} + {(3)}^{2}}

$c = \sqrt{{(4)}^{2} + {(3)}^{2}}$

Step 4

Raise

4

$4$ to the power of

2

$2$ .

c = \sqrt{16 + {(3)}^{2}}

$c = \sqrt{16 + {(3)}^{2}}$

Step 5

Raise

3

$3$ to the power of

2

$2$ .

c = \sqrt{16 + 9}

$c = \sqrt{16 + 9}$

Step 6

Add

16

$16$ and

9

$9$ .

c = \sqrt{25}

$c = \sqrt{25}$

Step 7

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

c = \sqrt{5^{2}}

$c = \sqrt{5^{2}}$

Step 8

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

c = 5

$c = 5$