Question

I need help on this question

Answer 1

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:

`CO=\sqrt[]{4^2-2^2}=\sqrt[]{14}`

And AO = 2

Final answer:

AB = 4

OB =4

OA = 4