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?

Guest Nov 11, 2022

#3**0 **

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

Guest Nov 11, 2022