Question

In a triangle, $ABC$ , if $\sum {\sin 3A} = 0$ , then it is.
$\left( A \right)Equilateral \\\ \left( B \right)Right{\text{ }}angled \\\ \left( C \right)Isosceles \\\ \left( D \right)Has{\text{ }}at{\text{ }}least{\text{ }}one{\text{ }}angle{\text{ }}{60^ \circ } \\\$

Answer 1

Hint : In the triangle $ABC$ , it is given that $\sum {\sin 3A} = 0$
Which means that, $\sin 3A + \sin 3B + \sin 3C = 0$
Now in order to state that which type of triangle it is, we need to simplify the expression, which can be done by using the formula for $\sin A + \sin B + \sin C = 0$
i.e. $\sin A + \sin B + \sin C = 4cos\left( {A/2} \right)cos\left( {B/2} \right)cos\left( {C/2} \right)$

Complete step-by-step answer :
First of all, we need to simplify the relation that is given, $\sum {\sin 3A} = 0$
$\sum {\sin 3A} = 0 \\\ \sin 3A + \sin 3B + \sin 3C = 0 \;$
Using the formula,
$\sin A + \sin B + \sin C = 4cos\left( {A/2} \right)cos\left( {B/2} \right)cos\left( {C/2} \right)$
In the above equation, we get,
$\Rightarrow 4\cos \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)cos\left( {\dfrac{C}{2}} \right) = 0 \\\ \Rightarrow \cos \left( {\dfrac{{3A}}{2}} \right)\cos \left( {\dfrac{{3B}}{2}} \right)cos\left( {\dfrac{{3C}}{2}} \right) = 0 \\\ \Rightarrow cos\left( {\dfrac{{3A}}{2}} \right) = 0 \\\ or \\\ cos\left( {\dfrac{{3B}}{2}} \right) = 0 \\\ or \\\ cos\left( {\dfrac{{3C}}{2}} \right) = 0 \;$
Taking
$cos\left( {\dfrac{{3A}}{2}} \right) = 0 \\\ \Rightarrow cos\left( {\dfrac{{3A}}{2}} \right) = cos\left( {\dfrac{\pi }{2}} \right) \\\ \Rightarrow \left( {\dfrac{{3A}}{2}} \right) = \left( {\dfrac{\pi }{2}} \right) \\\ \Rightarrow A = \dfrac{\pi }{3} \;$

Now if,
$cos\left( {\dfrac{{3B}}{2}} \right) = 0 \\\ \Rightarrow cos\left( {\dfrac{{3B}}{2}} \right) = cos\left( {\dfrac{\pi }{2}} \right) \\\ \Rightarrow \left( {\dfrac{{3B}}{2}} \right) = \left( {\dfrac{\pi }{2}} \right) \\\ \Rightarrow B = \dfrac{\pi }{3} \;$
Similarly,
$cos\left( {\dfrac{{3C}}{2}} \right) = 0 \\\ \Rightarrow cos\left( {\dfrac{{3C}}{2}} \right) = cos\left( {\dfrac{\pi }{2}} \right) \\\ \Rightarrow \left( {\dfrac{{3C}}{2}} \right) = \left( {\dfrac{\pi }{2}} \right) \\\ \Rightarrow C = \dfrac{\pi }{3} \;$
So, we can observe here that either $A$ , $B$ or $C$ .
Therefore, we have at least one angle in the triangle $ABC$ ,that equals ${60^ \circ }$ .
This answer matches the option $\left( D \right)Has{\text{ }}at{\text{ }}least{\text{ }}one{\text{ }}angle{\text{ }}{60^ \circ }$
However, we cannot say with surety that more than one angle will be equal to ${60^ \circ }$ only, it might be or might not be. So, we can’t say there will be two angles equal to ${60^ \circ }$ making it an isosceles triangle or three equal angles making it an equivalent angle. Also, from the given expression, it can’t be proved that any one angle is a right angle.
So, the only option that can be correct is (D)
So, the correct answer is “Option D”.

Note : The formula for sine functions addition, i.e. $\sin A + \sin B + \sin C = 4cos\left( {A/2} \right)cos\left( {B/2} \right)cos\left( {C/2} \right)$
And $\cos \dfrac{\pi }{2} = 0$ has also been used.
In the step,
$cos\left( {\dfrac{{3A}}{2}} \right) = 0 \\\ \Rightarrow cos\left( {\dfrac{{3A}}{2}} \right) = cos\left( {\dfrac{\pi }{2}} \right) \;$
Here we haven’t taken $\left( {2n + 1} \right)\dfrac{\pi }{2}$ , because all the angles are acute and, so the value of $n$ will always remain zero, and thus, we can directly write,
$\left( {\dfrac{{3C}}{2}} \right) = \left( {\dfrac{\pi }{2}} \right)$
Any of the three angles can be equal to ${60^ \circ }$ , so we can say that at least one of the angles is equal to ${60^ \circ }$ .