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

michaelcai  Nov 28, 2017

Best Answer 

 #1
avatar+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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1+0 Answers

 #1
avatar+1493 
+2
Best Answer

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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