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# Find Angles of a Triangle given side lengths

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The question is: The sides of a triangle are $$1, \sqrt{6}, \text{and } 1+\sqrt{3}$$

Find all the angles of the triangle.

I've tried using the Law of Cosines in reverse on the triangles to try and find the angles, but I just get a huge mess. Trying to calculate the angle between 1 and $$\sqrt{6}$$ I just get $$\arccos{(\frac{1}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{4})}$$ which clearly can't be right because that would just be a huge mess. Is there an easier way to do this I'm missing?

Nov 11, 2022

#1
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The anlges of the triangle are 20 degrees, 60 degrees, and 100 degrees.

Nov 11, 2022
#2
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That's obviously wrong

Guest Nov 11, 2022
#3
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Obtuse scalene triangle.

Sides: a = 1   b = 2.449   c = 2.732

Area: T = 1.219
Perimeter: p = 6.182
Semiperimeter: s = 3.091

Use the Law of Cosines to get:

Angle ∠ A =21.369° = 21°22'10″ = 0.373 rad
Angle ∠ B =63.195° = 63°11'40″ = 1.103 rad
Angle ∠ C =95.436° = 95°26'10″ = 1.666 rad

Nov 11, 2022