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# Find $AC$ in the following triangle, to two decimal places.​

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Find $AC$ in the following triangle, to two decimal places.

michaelcai  Nov 28, 2017

#1
+1493
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The length of $$\overline{AC}$$ can be found using the law of cosines. The law of cosines relates the remaining side by knowing two sides and its opposite included angle. I think this picture sums it up nicely.

Knowing this information, we can now find the missing length.

 $$c^2=a^2+b^2-2ab\cos C$$ Now, plug in the values we already know in the diagram and solve for the missing side. $$AC^2=1^2+3^2-2(1)(3)\cos40^{\circ}$$ Now, take the principal square root of both sides since the negative answer is nonsensical in the context of geometry. $$AC=\sqrt{1^2+3^2-2(1)(3)\cos40^{\circ}}\approx2.32$$ No units are given in the problem.
TheXSquaredFactor  Nov 28, 2017
edited by TheXSquaredFactor  Nov 28, 2017
edited by TheXSquaredFactor  Nov 28, 2017
Sort:

#1
+1493
+2

The length of $$\overline{AC}$$ can be found using the law of cosines. The law of cosines relates the remaining side by knowing two sides and its opposite included angle. I think this picture sums it up nicely.

Knowing this information, we can now find the missing length.

 $$c^2=a^2+b^2-2ab\cos C$$ Now, plug in the values we already know in the diagram and solve for the missing side. $$AC^2=1^2+3^2-2(1)(3)\cos40^{\circ}$$ Now, take the principal square root of both sides since the negative answer is nonsensical in the context of geometry. $$AC=\sqrt{1^2+3^2-2(1)(3)\cos40^{\circ}}\approx2.32$$ No units are given in the problem.
TheXSquaredFactor  Nov 28, 2017
edited by TheXSquaredFactor  Nov 28, 2017
edited by TheXSquaredFactor  Nov 28, 2017

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