Question

In a $ \vartriangle ABC $ , if $ \sin A\sin B = \dfrac{{ab}}{{{c^2}}} $ ,then the triangle is
1. equilateral
2. isosceles
3. right-angled
4. obtuse-angled

Answer 1

Hint: We use one of the concepts of geometry called sine rule in this problem. Law of sine or sine law states that in a triangle, the ratio of the side of a triangle to the sine of the opposite angle of that side is the same for all the sides of the triangle.
So according to the law of sines or sine law,
For any triangle ABC,

 $ \dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} $
Where $ a $ , $ b $ , $ c $ are the side lengths opposite to angles $ A $ , $ B $ , $ C $ of the triangle.

Complete step-by-step answer:
It is given in the question that,
 $ \sin A\sin B = \dfrac{{ab}}{{{c^2}}} $
On cross multiplication we get
 $ {c^2}\sin A\sin B = ab $
Taking sine terms on denominator of R.H.S , we get
 $ {c^2} = \dfrac{{ab}}{{\sin A\sin B}} $
Splitting the terms present in R.H.S we get,
 $ {c^2} = \left( {\dfrac{a}{{\sin A}}} \right)\left( {\dfrac{b}{{\sin B}}} \right) $
We know that from sine law or law of sine,
 $ \dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} $
Therefore the equation becomes,
 $ {c^2} = {\left( {\dfrac{c}{{\sin C}}} \right)^2} $
 $ \Rightarrow {c^2} = \dfrac{{{c^2}}}{{{{\sin }^2}C}} $
 $ \Rightarrow {\sin ^2}C = 1 $
Taking square root on both sides and simplifying we get,
 $ \Rightarrow \sin C = 1 $
We know that $ \sin {90^ \circ } = 1 $
 $ \therefore C = {90^ \circ } $
Hence the $ \vartriangle ABC $ is the right angled triangle.
So, the correct answer is “Option 3”.

Note: For a right angled triangle, one of the angles of the triangle must be right angle i.e $ {90^ \circ } $ . Hence the answer is a right angled triangle.
If it was an equilateral triangle then it should follow the condition of an equilateral triangle. We know that for the equilateral triangle, all the sides of the triangle must be of same length and all angles of the triangle must also be of equal measure.
If it was an isosceles triangle then it should be the condition of an isosceles triangle. We know that for an isosceles triangle, any two sides of a triangle must be of equal length and the angles opposite to equal sides of the triangle must also be equal in measure.