Find the coordinates of the vertices of an equilateral triangle of side 2a. Points are o(origin),A on x axis and Be above x axis
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One point is Origin.
The point on X-axis is given by:
the line makes an angle of 0 degrees with X-axis so,(0+2aCos0 ,0+2a sin0) ⇒ (2a,0)
The point above X-axis is given by:
this line makes 60 degrees with X-axis. So, point will be
(0+ 2aCos60, 0+2aSin60) ⇒ (a, √3 * a).
So, the points are (0,0) (2a,0), (a,√3 * a).
Is it understandable?
The point on X-axis is given by:
the line makes an angle of 0 degrees with X-axis so,(0+2aCos0 ,0+2a sin0) ⇒ (2a,0)
The point above X-axis is given by:
this line makes 60 degrees with X-axis. So, point will be
(0+ 2aCos60, 0+2aSin60) ⇒ (a, √3 * a).
So, the points are (0,0) (2a,0), (a,√3 * a).
Is it understandable?
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