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 \[{60^ \circ }\] .