Question

If b=3, c=4 and $\angle B = \dfrac{\pi }{3}$, then the number of triangles that can be constructed is
$
  (a){\text{ 0}} \\
  (b){\text{ 1}} \\
  (c){\text{ 3}} \\
  (d){\text{ 2}} \\
 $

Answer 1

Hint – In this question use the direct formula for sin rule that is $\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}$, where a, b and c are the sides and A, B and C are the angles. Substitute the values, this will help approaching the problem.

Complete step-by-step answer:
Given data
$b = 3,c = 4,B = \dfrac{\pi }{3}$
Now as we know that every triangle follows the rule of sin (i.e. the ratio of sin and their respective sides are always equal).
$ \Rightarrow \dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}$
$ \Rightarrow \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}$
Now substitute the values in this equation we have,
$ \Rightarrow \dfrac{{\sin \dfrac{\pi }{3}}}{3} = \dfrac{{\sin C}}{4}$
Now simplify the above equation we have,
$ \Rightarrow \sin C = \dfrac{4}{3} \times \sin \dfrac{\pi }{3} = \dfrac{4}{3} \times \dfrac{{\sqrt 3 }}{2} = \dfrac{{2\sqrt 3 }}{3}$, $\left[ {\because \sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}} \right]$
Now as we know $ - 1 \leqslant \sin \theta \leqslant 1$
But $\dfrac{{2\sqrt 3 }}{3} > 1$
$ \Rightarrow \sin C > 1$ (which is not possible)
Hence, no triangle is possible.
So this is the required answer.
Hence option (A) is correct.

Note – The key trick involved here was regarding the range of sin function. Sin is a periodic function whose range belongs to the interval of [-1,1] and period is $2\pi $. So the value of sin can’t exceed 1 and can be lowered than -1, that’s why no triangle is possible for the angles obtained as $\dfrac{{2\sqrt 3 }}{3} > 1$.