Solution
For this case we know that m < AOB = 60
BC =2
And we need to find AB, OB and OA
From the picture we also know that:
m < ACO = m < BCO =90º
Since triangles ACO and BCO are similar we can conclude that:
m < AOC = m< BOC = 30º
Then we have that:
m Then we can find
cos 60 = 2/OB
OB = 2/cos60 = 4
And using pythagoras we got:
[tex]CO=\sqrt[]{4^2-2^2}=\sqrt[]{14}[/tex]And AO = 2
Final answer:
AB = 4
OB =4
OA = 4
