Consider a triangle ΔABC, where sides AB, BC and angle ∡C are given.
Angle ∡C is the angle opposite of side AB.
It says that the given side opposite the given angle is less than the other given side. It means AB < BC.
It says that the ratio of the longer side to the shorter side, multiplied by the sine of the angle opposite the shorter side, is less than 1. It means
[tex]\frac{BC}{AB}[/tex] · sin(C) < 1
We know about Law of Sines of Triangle is given by :-
[tex]\frac{AB}{sin(C)} =\frac{BC}{sin(A)} =\frac{AC}{sin(B)}[/tex]
[tex]< br/ > Solving[/tex] [tex]\frac{AB}{sin(C)} =\frac{BC}{sin(A)}[/tex]
[tex]< br/ > Cross[/tex] [tex]Multiplying[/tex]
[tex]< br/ > AB[/tex] [tex]sin(A)=BC[/tex] [tex]sin(C)[/tex]
[tex]< br/ > sin(A)=\frac{BC}{AB}[/tex] [tex]sin(C)[/tex]
[tex]< br/ > sin(A) < 1[/tex] \\
We got sin(A) < 1, and BC > AB.
Therefore, angle ∡A could be either acute or obtuse angle.
So, There will be two solutions for the angle ∡A.
⇒ option D is the final answer.
