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In triangle ABC, a=6.3. B=9.4, <C=38.3° how do you find side C?

1 Answer

May 10, 2018

The length of side $c$ $c$ is approximately 5.9.

Explanation:

You use the Law of Cosines - 'cause it's the LAW!

$c^{2} = a^{2} + b^{2} - 2 a b cos θ$ $c^2=a^2+b^2-2abcostheta$

Soving for $c$ $c$ gives us the equation

$c = \sqrt{a^{2} + b^{2} - 2 a b cos θ}$ $c=sqrt(a^2+b^2-2abcostheta)$

Here $a = 6.3$ $a=6.3$ , $b = 9.4$ $b=9.4$ , and $θ = {38.3}^{\circ}$ $theta=38.3^@$ , so

$c = \sqrt{{6.3}^{2} + {9.4}^{2} - 2 (6.3) (9.4) {cos 38.3}^{\circ}} \approx 5.9$ $c=sqrt(6.3^2+9.4^2-2(6.3)(9.4)cos38.3^@)~~5.9$

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