Question

If $b=3,c=4$ and $B=\dfrac{\pi }{3}$ then the number of the triangle that can be constructed is
A. Infinite
B. Two
C. One
D. Nil

Answer 1

Hint: We will use Law of sine and take only angles $B,C$ and sides \[b,c\]. Then we will substitute the value of angles and sides given and derive the value of angle $C$. We will then check the value of angle $C$ and see if it lies in the interval of sine or not. On the basis of that we can determine how many triangles can be formed.
The value of sine lies in the interval of $-1\le \sin \theta \le 1$.

Complete step by step solution: We are given sides of a triangle $b=3,c=4$ and one angle $B=\dfrac{\pi }{3}$ and we have to determine how many triangles can be formed with this.
By using Law of sines $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$, we will take,
$\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
We will now substitute the given values in the above equation.
$\begin{align}
  & \dfrac{3}{\sin \left( \dfrac{\pi }{3} \right)}=\dfrac{4}{\sin C}\,\,\, \\
 & \dfrac{3}{\dfrac{\sqrt{3}}{2}}=\dfrac{4}{\sin C} \\
 & \sin C=\dfrac{4\times \dfrac{\sqrt{3}}{2}}{3} \\
 & \sin C=\dfrac{2}{\sqrt{3}}
\end{align}$
The value of the $\sin C$ we derived is greater than one, which lies outside the interval of the sine. Hence this value of the angle $C$ of the triangle is not possible. Hence no triangle will be possible.

The number of triangles which can be constructed with sides of length $b=3,c=4$ and angle $B=\dfrac{\pi }{3}$ is nil so that no triangles can be possibly constructed. Hence the correct option is (D).

Note: The Law of sines holds true for any triangle. It shows the relationship between the ratio of the length of the sides to their opposite angles. Each of the trigonometric angles has an interval in which their value lies. The interval of the sine angle lies in $\left[ -\pi ,\pi \right]$.