Trigonometry Examples

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\begin{matrix} Side & Angle \begin{matrix} b = c = & 9 a = \end{matrix} & \begin{matrix} A = & 45 B = & 45 C = & 90 \end{matrix} \end{matrix}

$\begin{matrix} Side & Angle \\ \begin{matrix} b = \\ c = & 9 \\ a = \end{matrix} & \begin{matrix} A = & 45 \\ B = & 45 \\ C = & 90 \end{matrix} \end{matrix}$

Step 1

Find

b

$b$ .

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Step 1.1

The cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse.

cos (A) = \frac{adj}{hyp}

$\cos (A) = \frac{adj}{hyp}$

Step 1.2

Substitute the name of each side into the definition of the cosine function.

cos (A) = \frac{b}{c}

$\cos (A) = \frac{b}{c}$

Step 1.3

Set up the equation to solve for the adjacent side, in this case

b

$b$ .

b = c \cdot cos (A)

$b = c \cdot \cos (A)$

Step 1.4

Substitute the values of each variable into the formula for cosine.

b = 9 \cdot cos (45)

$b = 9 \cdot \cos (45)$

Step 1.5

Combine

9

$9$ and

\frac{\sqrt{2}}{2}

$\frac{\sqrt{2}}{2}$ .

b = \frac{9 \sqrt{2}}{2}

$b = \frac{9 \sqrt{2}}{2}$

b = \frac{9 \sqrt{2}}{2}

$b = \frac{9 \sqrt{2}}{2}$

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

a

$a$ .

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

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

Step 2.3

Substitute the actual values into the equation.

a = \sqrt{{(9)}^{2} - {(\frac{9 \sqrt{2}}{2})}^{2}}

$a = \sqrt{{(9)}^{2} - {(\frac{9 \sqrt{2}}{2})}^{2}}$

Step 2.4

Raise

9

$9$ to the power of

2

$2$ .

a = \sqrt{81 - {(\frac{9 \sqrt{2}}{2})}^{2}}

$a = \sqrt{81 - {(\frac{9 \sqrt{2}}{2})}^{2}}$

Step 2.5

Use the power rule

{(a b)}^{n} = a^{n} b^{n}

${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 2.5.1

Apply the product rule to

\frac{9 \sqrt{2}}{2}

$\frac{9 \sqrt{2}}{2}$ .

a = \sqrt{81 - \frac{{(9 \sqrt{2})}^{2}}{2^{2}}}

$a = \sqrt{81 - \frac{{(9 \sqrt{2})}^{2}}{2^{2}}}$

Step 2.5.2

Apply the product rule to

9 \sqrt{2}

$9 \sqrt{2}$ .

a = \sqrt{81 - \frac{9^{2} {\sqrt{2}}^{2}}{2^{2}}}

$a = \sqrt{81 - \frac{9^{2} {\sqrt{2}}^{2}}{2^{2}}}$

a = \sqrt{81 - \frac{9^{2} {\sqrt{2}}^{2}}{2^{2}}}

$a = \sqrt{81 - \frac{9^{2} {\sqrt{2}}^{2}}{2^{2}}}$

Step 2.6

Simplify the numerator.

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Step 2.6.1

Raise

9

$9$ to the power of

2

$2$ .

a = \sqrt{81 - \frac{81 {\sqrt{2}}^{2}}{2^{2}}}

$a = \sqrt{81 - \frac{81 {\sqrt{2}}^{2}}{2^{2}}}$

Step 2.6.2

Rewrite

{\sqrt{2}}^{2}

${\sqrt{2}}^{2}$ as

2

$2$ .

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Step 2.6.2.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt{2}

$\sqrt{2}$ as

2^{\frac{1}{2}}

$2^{\frac{1}{2}}$ .

a = \sqrt{81 - \frac{81 {(2^{\frac{1}{2}})}^{2}}{2^{2}}}

$a = \sqrt{81 - \frac{81 {(2^{\frac{1}{2}})}^{2}}{2^{2}}}$

Step 2.6.2.2

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

a = \sqrt{81 - \frac{81 \cdot 2^{\frac{1}{2} \cdot 2}}{2^{2}}}

$a = \sqrt{81 - \frac{81 \cdot 2^{\frac{1}{2} \cdot 2}}{2^{2}}}$

Step 2.6.2.3

Combine

\frac{1}{2}

$\frac{1}{2}$ and

2

$2$ .

a = \sqrt{81 - \frac{81 \cdot 2^{\frac{2}{2}}}{2^{2}}}

$a = \sqrt{81 - \frac{81 \cdot 2^{\frac{2}{2}}}{2^{2}}}$

Step 2.6.2.4

Cancel the common factor of

2

$2$ .

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Step 2.6.2.4.1

Cancel the common factor.

a = \sqrt{81 - \frac{81 \cdot 2^{\frac{2}{2}}}{2^{2}}}

$a = \sqrt{81 - \frac{81 \cdot 2^{\frac{2}{2}}}{2^{2}}}$

Step 2.6.2.4.2

Rewrite the expression.

a = \sqrt{81 - \frac{81 \cdot 2}{2^{2}}}