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Let z1;z2&z3 represent the vertices of A ;B & C of the triangle ABC respectively in the argand plane such that |z1|=|z2|=|z3|=5 prove that z1sin2A+z2sin2B+z3sin2C=0
Let z1;z2&z3 represent the vertices of A ;B & C of the triangle ABC respectively in the argand plane such that |z1|=|z2|=|z3|=5 prove that z1sin2A+z2sin2B+z3sin2C=0

```
3 years ago

```							If A-z1, B-z2 and C-z3 are vertices of the triangle, then its centroid G will be (z1+z2+z3)/3Now, |z1|=|z2|=|z3| => the points A,B,C lie on a circle with center as the origin O.Therefore, O is the circumcenter of the triangle.But for an equilateral triangle, the circumcenter coincides with the centroid.Therefore, the centroid of the triangle is Oor(z1+z2+z3)/3 = 0or (z1+z2+z3) = 0Now for an equilateral traingle,Angle A =angle B= angle C =60°Thus We can write the question as Sin2A(z1 + z2 + z3)But we know that(z1 + z2 + z3) =0Hencez1sin2A+z2sin2B+z3sin2C=0
```
3 years ago
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