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Question

If the three points (a,b),(a+rcosα,b+rsinα),(a+rcosβ,b+rsinβ) are the vertices of an equilateral triangle. Find the value of |αβ|

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Solution

Let A(a,b);B(a+rcosα,b+sinα);C(a+rcosβ,b+rsinβ)

We have
AC=(a+rcosβa)2+(b+rsinβb)2

AC=(r2cos2β)+(r2sin2β)=r2=r

AC=r

Now,

BC=(a+rcosαarcosβ)2+(b+rsinαbrsinβ)2

BC=r2(cosαcosβ)2+r2(sinαsinβ)2

BC=r(cos2α+cos2β2cosαcosβ)+(sin2α+sin2β2sinαsinβ)

BC=r1+12cosαcosβ2sinαsinβ

BC=r22(cosαcosβ+sinαsinβ)

BC=r22cos(αβ)

As ABC is equilateral triangle, AC=BC

r22cos(αβ)=r

22cos(αβ)=1

22cos(αβ)=1

cos(αβ)=12

Therefore, |αβ|=π3

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