Question

Find the coordinates of the vertices of an equilateral triangle of side 2a as shown in Fig 14.5

Answer 1

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
${OB}^2={OL}^2+{LB}^2$
$\Rightarrow {\left(2a\right)}^2=a^2+{LB}^2$
$\Rightarrow {LB}^2={3a}^2$
$\Rightarrow LB=\sqrt{3}a$
Clearly, coordinates of O are (0,0) and that of A are (2a,0). Since OL=a and $LB=\sqrt{3}a$. So, the coordinates of B are $(a,\sqrt{3}a).$