Question 1101717
Draw this and let angle a be opposite to 2, angle b opposite to side sqrt(6) and angle c be opposite to side (1+ sqrt(3)).
Law of cosines
a^2=b^2+c^2-2bc cos A
4=6+(2.732)^2-2(sqrt (6)*(1+sqrt(3)) cos A
numerically
4=6+7.46-2*6.69 cos A
-9.46=-13.38 cos A
cos A=9.46/13.38
arc cos both
A=45.0 degrees.
find b
b^2=a^2+c^2-2ac cos B
6=4+7.46-2(2*2.732) cos B
-5.46=-10.93 cos B
cos B=0.4995
B=60.0 degrees.
find c (besides subtracting 45 and 60 from 180)
c^2=a^2+b^2-2ab cos C
7.46=4+6-2(2 sqrt(6)) cos C
-2.54=-4 sqrt(6) cos C
-2.54=-9.80 cos C
cos C=2.54/9.80=75.0 deg (rounding from 74.97)
The ratios are 45:60:75 or 3:4:5