How do you solve the right triangle ABC given a=2, c=7?

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How do you solve the right triangle ABC given a=2, c=7?

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Answer 1 · 2017-01-02T20:34:26

$b = 3 \sqrt{5}$ $b=3sqrt(5)$ (assuming the side with length $c$ $c$ is the hypotenuse)

Explanation:

If $c$ $c$ is the hypotenuse then
$XXX c^{2} = a^{2} + b^{2}$ $color(white)("XXX")c^2=a^2+b^2$
or
$XXX b^{2} = c^{2} - a^{2}$ $color(white)("XXX")b^2=c^2-a^2$

for the given values of $a$ $a$ and $c$ $c$
$XXX b^{2} = 7^{2} - 2^{2} = 49 - 4 = 45$ $color(white)("XXX")b^2=7^2-2^2=49-4=45$
which implies
$XXX b = \sqrt{45} = \sqrt{3^{2} \cdot 5} = 3 \sqrt{5}$ $color(white)("XXX")b=sqrt(45)=sqrt(3^2 * 5) = 3sqrt(5)$

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If the side with length $b$ $b$ is the hypotenuse then
(using logic similar to the above)
$XXX b = = \sqrt{53}$ $color(white)("XXX")b==sqrt(53)$

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How do you solve the right triangle ABC given a=2, c=7?

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