How do you solve the triangle given $m\angle B={{105}^{\circ }},b=23,a=14$ ?

Hint: We solve this problem using the Lami’s formula which is $\dfrac{a}{\sin \angle A}=\dfrac{b}{\sin \angle B}=\dfrac{c}{\sin \angle C}$ . We divide the formula into two separate formulae and then put the known values to get the desired values.

Complete step by step solution:
The details that we have been provided about the triangle $\Delta ABC$ are that
$m\angle B={{105}^{\circ }},b=23,a=14$
We can see that as we are given the length of a side and its opposite angle, we can easily apply the Lami’s theorem. Lami's theorem states that for any triangle $\Delta ABC$ , if $a,b,c$ are the three sides of the triangle and $\angle A,\angle B,\angle C$ are the three angles, then there is a formula which relates all of them. This formula is,
$\dfrac{a}{\sin \angle A}=\dfrac{b}{\sin \angle B}=\dfrac{c}{\sin \angle C}$
This huge equation can be broken down into two equations, which are
$\dfrac{a}{\sin \angle A}=\dfrac{b}{\sin \angle B}....\left( 1 \right)$
$\dfrac{c}{\sin \angle C}=\dfrac{b}{\sin \angle B}....\left( 2 \right)$
Let us first solve the equation $\left( 1 \right)$ . We put the values of $a,b,\angle B$ in the equation to get,
$\Rightarrow \dfrac{14}{\sin \angle A}=\dfrac{23}{\sin \angle {{105}^{\circ }}}$
Taking reciprocals on both sides of the above equation, we get,
$\Rightarrow \dfrac{\sin \angle A}{14}=\dfrac{\sin \angle {{105}^{\circ }}}{23}$
Multiplying both sides of the above equation by $14$ , we get,
$\Rightarrow \sin \angle A=\dfrac{\sin \angle {{105}^{\circ }}}{23}\times 14$
Upon simplification, we get,
$\Rightarrow \sin \angle A=\dfrac{0.9659}{1.6428}$
This gives,
$\Rightarrow \sin \angle A=0.5879$
Taking ${{\sin }^{-1}}$ on both sides, we get,
$\begin{align}
  & \Rightarrow \angle A={{\sin }^{-1}}\left( 0.5879 \right) \\
 & \Rightarrow \angle A={{36.012}^{\circ }} \\
\end{align}$
Let us now solve the equation $\left( 2 \right)$ . But, there are two unknowns and only one equation available. Thus, we have to incorporate a second equation, which is the sum of internal angles of a triangle,
$\angle A+\angle B+\angle C={{180}^{\circ }}$
Putting the values of $\angle A,\angle B$ in the above equation, we get,
$\begin{align}
  & \Rightarrow {{36.012}^{\circ }}+{{105}^{\circ }}+\angle C={{180}^{\circ }} \\
 & \Rightarrow \angle C={{38.988}^{\circ }} \\
\end{align}$
We can now solve the equation $\left( 2 \right)$ easily. Putting the values of $\angle C,\angle B,b$ ,we get,
$\begin{align}
  & \Rightarrow \dfrac{c}{\sin {{38.988}^{\circ }}}=\dfrac{23}{\sin {{105}^{\circ }}} \\
 & \Rightarrow c=\dfrac{23}{\sin {{105}^{\circ }}}\times \sin {{38.988}^{\circ }} \\
 & \Rightarrow c=14.981 \\
\end{align}$
Therefore, we can conclude that the triangle $\Delta ABC$ has the angles $\angle A={{36.012}^{\circ }},\angle B={{105}^{\circ }},\angle C={{38.988}^{\circ }}$ and the sides $a=14,b=23,c=14.981$.


Note: We should apply the Lami’s theorem properly and should remember that the formula is only valid for $\text{sine}$ trigonometric ratio and not for any other trigonometric ratios. Most of the students make mistakes here. Also, it is not advisable to use this formula for more than three variables.