Find the coordinates of the vertices of an equilateral triangle of side 2a which lies in the first quadrant and one of it's vertices is (0,0)
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Since OAB is an equilateral triangle of side 2a. Therefore,
OA=AB=OB=2a
Let BL perpendicular from B on OA. Then,
OL=LA+a
In △OLB, we have
OB 2
=OL 2 +LB 2
⇒(2a) 2 =a 2 +LB 2
⇒LB 2 =3a 2
⇒LB= 3 a
Clearly, coordinates of O are (0,0) and that of A are (2a,0). Since OL=a and LB= 3
a. So, the coordinates of B are (a, 3a).
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