Prove that the points 2a-4a- 2a-6a and 2a-3a-5a are the vertices of an equilateral triangle-

PROOF : AB=√( (2a 2a)^2 + (6a 4a)^2 ) =2a AC=√(( 2a ( 2a+√( 3 ) a ) #160; ) #160; ^ 2 +( 4a 5a ) #160; ^ 2 ) =√( 3 a ^ 2 + a ^ 2 ) ...

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Shortest Distance between Two Skew Lines

Prove that th...

Question

Prove that the points $(2 a, 4 a), (2 a, 6 a)$ and $(2 a + \sqrt{3} a, 5 a)$ are the vertices of an equilateral triangle.

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(2 a, 4 a), (2 a, 6 a)

(2 a + \sqrt{3} a, 5 a)

(2 a, 4 a)

(2 a, 6 a)

(2 a + \sqrt{3} a, 5 a)

a > 0

(2 a, 4 a), (2 a, 6 a)

(2 a + \sqrt{3} a, 5 a)

(2 a + a \sqrt{3}, 5 a) .

\frac{cos 2 A cos 3 A - cos 2 A cos 7 A + cos A cos 10 A}{sin 4 A sin 3 A - sin 2 A sin 5 A + sin 4 A sin 7 A} = cot 6 A cot 5 A

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