Question

In a triangle ABC, if \[{\sin ^2}A + {\sin ^2}B + {\sin ^2}C = 2\] then the triangle is
a) right angled triangle but need not be isosceles
b) right angled and isosceles
c) isosceles but need not be right angled
d) equilateral

Answer 1

Hint: Here we have to determine the type of the triangle. The given equation is in terms of trigonometric ratios. So we use the trigonometric identities, trigonometric functions for the allied angles and trigonometric sum or difference formulas to determine the angle of a triangle.

Complete answer:
Now consider the given inequality
\[ \Rightarrow {\sin ^2}A + {\sin ^2}B + {\sin ^2}C = 2\]
The number 2 can be written as sum of 1 and 1
\[ \Rightarrow {\sin ^2}A + {\sin ^2}B + {\sin ^2}C = 1 + 1\]
Take the terms which are present in LHS to the RHS.
\[ \Rightarrow 1 - {\sin ^2}A + 1 - {\sin ^2}B - {\sin ^2}C = 0\]
By the trigonometric identities we know that \[{\sin ^2}x + {\cos ^2}x = 1\, \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x\], so the above inequality is written as
\[ \Rightarrow {\cos ^2}A + {\cos ^2}B - {\sin ^2}C = 0\]
On multiplying by 2
\[ \Rightarrow 2{\cos ^2}A + 2{\cos ^2}B - 2{\sin ^2}C = 0\]
By the trigonometric identities we know that \[{\sin ^2}x + {\cos ^2}x = 1\, \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x\], so the above inequality is written as
\[ \Rightarrow 2{\cos ^2}A + 2{\cos ^2}B - 2(1 - {\cos ^2}C) = 0\]
\[ \Rightarrow 2{\cos ^2}A + 2{\cos ^2}B - 2 + 2{\cos ^2}C = 0\]
As we know that \[\cos 2A = 2{\cos ^2}A - 1\], so first two terms we rewrite using this formula.
\[ \Rightarrow 1 + \cos 2A + 1 + \cos 2B - 2 + 2{\cos ^2}C = 0\]
On simplifying we have
\[ \Rightarrow \cos 2A + \cos 2B + 2{\cos ^2}C = 0\]
By the trigonometric formula we have \[\cos 2A + \cos 2B = 2\cos (A + B)\cos (A - B)\], so we have
\[ \Rightarrow 2\cos (A + B)\cos (A - B) + 2{\cos ^2}C = 0\]
Take 2 as common from both the terms and cancel it we have
\[ \Rightarrow \cos (A + B)\cos (A - B) + {\cos ^2}C = 0\]
\[ \Rightarrow \cos (A + B)\cos (A - B) + \cos C.\cos C = 0\]
As we know that the ABC is a triangle. Sum of all angles in triangle is equal to 180 degrees so we have (\[(A + B + C) = \pi \])
Now the above equation can be written as
\[ \Rightarrow \cos (\pi - C)\cos (A - B) + \cos C.\cos (\pi - (A + B)) = 0\]
On considering the trigonometric function for allied angles.
\[ \Rightarrow - \cos C\cos (A - B) - \cos C.\cos (A + B) = 0\]
On multiplying the above equation by -1.
\[ \Rightarrow \cos C\cos (A - B) + \cos C.\cos (A + B) = 0\]
Take \[\cos C\] as common we have
\[ \Rightarrow \cos C(\cos (A - B) + \cos (A + B)) = 0\]
On simplifying we have
\[ \Rightarrow \cos C(2\cos A\cos B) = 0\]
\[ \Rightarrow \cos C = 0\] or \[\cos B = 0\] or \[\cos A = 0\]
Therefore this implies that any one of the angles is \[{90^ \circ }\].
Let us assume that the angle A = \[{90^ \circ }\] and the sum of the other two angles will be \[{90^ \circ }\]. Therefore \[B + C = \dfrac{\pi }{2}\].
Now consider given question
\[ \Rightarrow {\sin ^2}\left( {\dfrac{\pi }{2}} \right) + {\sin ^2}\left( {\dfrac{\pi }{2} - C} \right) + {\sin ^2}C = 2\]
On considering the trigonometric function for allied angles.
\[ \Rightarrow {\sin ^2}\left( {\dfrac{\pi }{2}} \right) + {\cos ^2}C + {\sin ^2}C = 2\]
By considering trigonometric identities and tables of trigonometric functions we get
\[ \Rightarrow 1 + 1 = 2\]
Therefore the given triangle is a right angled triangle and the sum of two sides of a triangle is equal to \[{90^ \circ }\], so it need not be an isosceles.
Hence option A is the correct one.

Note:
In a triangle we have three kinds of triangles namely, equilateral triangle, isosceles triangle and scalene triangle. In an equilateral triangle all the angles are equal. In an isosceles triangle any two angles are equal. In the scalene triangle all angles are different. In this question the sum of two angles is equal to \[{90^ \circ }\], but it is not mentioned that both angles are equal in degree. So we can say that the triangle need not be isosceles.