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Question

In Δ ABC having vertices A(a cosθ1,a sinθ1),B(a cosθ2,a sinθ2) and C(a cosθ3,a sinθ3) is equilateral, then which of the followings is/are true?

A
cosθ1+cosθ2+cosθ3=0
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B
sinθ1+sinθ2+sinθ3=0
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C
cos(θ1θ2)+cos(θ2θ3)+cos(θ3θ1)=32
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D
sin(θ1θ22)=sin(θ2θ32)=sin(θ1θ32)
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Solution

The correct options are
A cosθ1+cosθ2+cosθ3=0
B sinθ1+sinθ2+sinθ3=0
C cos(θ1θ2)+cos(θ2θ3)+cos(θ3θ1)=32
D sin(θ1θ22)=sin(θ2θ32)=sin(θ1θ32)
Here, notice that the vertices of the triangle ABC are:
A (a cosθ1,asinθ1)B (a cosθ2,a sinθ2)C (a cosθ3,a sinθ3)

These are similar to the polar form of a point in complex numbers implicating that all 3 vertices of ΔABC are at a distance of `a’ units from the origin and only differing in their arguments.
So, all there vertices lie on a circle of radius `a’ and centre as origin which makes the circumcentre of the triangle ABC as the origin.

ΔABC being an equilateral triangle; the circumcentre will be the same as the centroid.
CentroidG(0,0)::(a cosθ1+a cosθ2+a cos θ33a sinθ1+a sinθ2+a sinθ32)
cosθ1+cosθ2+cosθ3=0 option(a)
& sinθ1+sinθ2+sinθ3=0 option (b)
Now, squaring & adding these two, we get
(cosθ1+cosθ2+cosθ3)2+(sinθ1+sinθ2+sinθ3)2=0
1+1+1+2{(cosθ1cosθ2+sinθ1sinθ2)+(cosθ2cosθ3+sinθ2sinθ3)+(cosθ1cosθ3+sinθ1sinθ3)}=0
cosθ1θ2)+cos(θ2θ3)+cos(θ3θ1)=32 .....option (c)

In equilateral triangle; AB = BC = AC
AB=(a cosθ2a cos θ1)2+(a sinθ2a sinθ1)2
=a2{1+12(sinθ1sinθ2+cosθ1cosθ2)}
=a22cos(θ1θ2)
=2asin(θ1θ22)
Similarly: BC=2asin(θ3θ12)
sin(θ1θ22)=sin(θ2θ32)=sin(θ3θ12) ....option (d)


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