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How do you solve the right triangle ABC given Side a = 8, Side b = 10, Side c = 5?

1 Answer

Jun 4, 2015

$\sqrt{8^{2} + 5^{2}} \neq \sqrt{10^{2}}$ $sqrt(8^2+5^2) != sqrt(10^2)$

Therefore
$XXXX$ $color(white)("XXXX")$ a triangle with sides $a = 8$ $a=8$ , $b = 10$ $b=10$ , and $c = 5$ $c=5$ is not a right triangle.

With the given sides we can determine the angles using the Law of Cosines. For example
$XXXX$ $color(white)("XXXX")$ $cos (C) = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$ $cos(C) = (a^2+b^2-c^2)/(2ab)$

$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $= \frac{64 + 100 - 25}{160} = \frac{139}{160}$ $= (64+100-25)/(160) = 139/160$

$XXXX$ $color(white)("XXXX")$ $C = arccos (\frac{139}{160})$ $C = arccos(139/160)$

$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $= 0.518124 radians = {29.6863}^{\circ}$ $= 0.518124 " radians " = 29.6863^@$

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