Question

In $\Delta$ ABC having vertices $A(acos{\theta }_1,asin{\theta }_1),B(acos{\theta }_2,asin{\theta }_2)$ and $C(acos{\theta }_3,asin{\theta }_3)$ is equilateral, then which of the followings is/are true?

• A. $cos{\theta }_1+cos{\theta }_2+cos{\theta }_3=0$
• B. $sin{\theta }_1+sin{\theta }_2+sin{\theta }_3=0$
• C. $cos({\theta }_1-{\theta }_2)+cos({\theta }_2-{\theta }_3)+cos({\theta }_3-{\theta }_1)=\frac{-3}{2}$
• D. $\left|sin\left(\frac{{\theta }_1-{\theta }_2}{2}\right)\right|=sin\left|\left(\frac{{\theta }_2-{\theta }_3}{2}\right)\right|=\left|sin\left(\frac{{\theta }_1-{\theta }_3}{2}\right)\right|$

Answer 1

Answer: (A), (B), (C), (D)

$\left|sin\left(\frac{{\theta }_1-{\theta }_2}{2}\right)\right|=sin\left|\left(\frac{{\theta }_2-{\theta }_3}{2}\right)\right|=\left|sin\left(\frac{{\theta }_1-{\theta }_3}{2}\right)\right|$

Here, notice that the vertices of the triangle ABC are:
$A(acos{\theta }_1,asin{\theta }_1)B(acos{\theta }_2,asin{\theta }_2)C(acos{\theta }_3,asin{\theta }_3)$

These are similar to the polar form of a point in complex numbers implicating that all 3 vertices of $\Delta ABC$ 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.

$\Delta ABC$ being an equilateral triangle; the circumcentre will be the same as the centroid.
$\therefore CentroidG(0,0)::\left(\frac{acos{\theta }_1+acos{\theta }_2+acos{\theta }_3}{3}\frac{asin{\theta }_1+asin{\theta }_2+asin{\theta }_3}{2}\right)$
$\Rightarrow cos{\theta }_1+cos{\theta }_2+cos{\theta }_3=0$ option(a)
$\&sin{\theta }_1+sin{\theta }_2+sin{\theta }_3=0$ option (b)
Now, squaring & adding these two, we get
$(cos{\theta }_1+cos{\theta }_2+cos{\theta }_3{)}^2+(sin{\theta }_1+sin{\theta }_2+sin{\theta }_3{)}^2=0$
$\Rightarrow 1+1+1+2\left\{\right(cos{\theta }_{1}cos{\theta }_2+sin{\theta }_{1}sin{\theta }_2)+(cos{\theta }_{2}cos{\theta }_3+sin{\theta }_{2}sin{\theta }_3)+(cos{\theta }_{1}cos{\theta }_3+sin{\theta }_{1}sin{\theta }_3\left)\right\}=0$
$\Rightarrow cos{\theta }_1-{\theta }_2)+cos({\theta }_2-{\theta }_3)+cos({\theta }_3-{\theta }_1)=-\frac{3}{2}$ .....option (c)

In equilateral triangle; AB = BC = AC
$AB=\sqrt{(acos{\theta }_2-acos{\theta }_1{)}^2+(asin{\theta }_2-asin{\theta }_1{)}^2}$
$=\sqrt{a^2\{1+1-2(sin{\theta }_{1}sin{\theta }_2+cos{\theta }_{1}cos{\theta }_2\left)\right\}}$
$=a\sqrt{2-2cos({\theta }_1-{\theta }_2)}$
$=2a\left|sin\left(\frac{{\theta }_1-{\theta }_2}{2}\right)\right|$
Similarly: $BC=2a\left|sin\left(\frac{{\theta }_3-{\theta }_1}{2}\right)\right|$
$\therefore \left|sin\left(\frac{{\theta }_1-{\theta }_2}{2}\right)\right|=sin\left|\left(\frac{{\theta }_2-{\theta }_3}{2}\right)\right|=\left|sin\left(\frac{{\theta }_3-{\theta }_1}{2}\right)\right|$ ....option (d)