Let triangle ABC be an equilateral triangle. There is a point O insie the triangle such that angle AOB = 115 degrees and angle BOC = 125 degrees. Ir a triangle is constructed with sides with length AO, BO, and CO, find the degree measure of the largest interior angle of this triangle.
I have not been able to work it out algebraically.
So I estimated it graphically. (using a free program GeoGebra) [Not what you want, I know this]
The stupid pic has deleted itself so I cannot show it to you. bummer!
(Geogebra is playing up - this has not happened to me before)
The sides are approximately in the ratio 65:62:59
The biggest angle is opposite the longest side.
Using cosine rule I got the biggest angle as approx 64.9 degrees.
Anyway. I am sure there is some sensible algebraic way to do it.
Turns out that the largest angle is 66.468 (3 dp ) deg.
Call the angle OAC alpha say, and run that round the triangle,
OAB = 60 - alpha, OBA = 180 - 115 - (60 - alpha), and so on.
Calling AO x, BO y and CO z, then use the sin rule twice, in triangles AOB and AOC,
equate the two to derive an expression for tan (alpha) - I got alpha = 28.3296 deg.
From that, using the sin rule in the three triangles, calculate x, y an z.
I got (to 4 dp) x = 0.6063, y = 0.5793, z = 0.5236.
Then the cosine rule to calculate the largest angle.