Find the Length of a tri{12}{}{13}{}{}{}

 $\begin{matrix} Side & Angle \\ \begin{matrix} b = & 12 \\ c = & 13 \\ a = \end{matrix} & \begin{matrix} A = \\ B = \\ C = \end{matrix} \end{matrix}$

Assume that angle $C = 90$ .

$C = 90$

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

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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).

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

Solve the equation for $a$ .

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

Substitute the actual values into the equation.

$a = \sqrt{{(13)}^{2} - {(12)}^{2}}$

Raise $13$ to the power of $2$ .

$a = \sqrt{169 - {(12)}^{2}}$

Raise $12$ to the power of $2$ .

$a = \sqrt{169 - 1 \cdot 144}$

Multiply $- 1$ by $144$ .

$a = \sqrt{169 - 144}$

Subtract $144$ from $169$ .

$a = \sqrt{25}$

Rewrite $25$ as $5^{2}$ .

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

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

$a = 5$

Find $B$ .

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

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

Substitute in the values of the opposite side to angle $B$ and hypotenuse $13$ of the triangle.

$B = \arcsin (\frac{12}{13})$

Evaluate $\arcsin (\frac{12}{13})$ .

$B = 67.38013505$

Find the last angle of the triangle.

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The sum of all the angles in a triangle is $180$ degrees.

$A + 90 + 67.38013505 = 180$

Solve the equation for $A$ .

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Add $90$ and $67.38013505$ .

$A + 157.38013505 = 180$

Move all terms not containing $A$ to the right side of the equation.

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Subtract $157.38013505$ from both sides of the equation.

$A = 180 - 157.38013505$

Subtract $157.38013505$ from $180$ .

$A = 22.61986494$

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

$A = 22.61986494$

$B = 67.38013505$

$C = 90$

$a = 5$

$b = 12$

$c = 13$

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