Question

If the area of a triangle $ABC$ is $\Delta $, then ${{a}^{2}}\sin 2B+{{b}^{2}}\sin 2A$ is equal to
A. $3\Delta $
B. $2\Delta $
C. $4\Delta $
D. $5\Delta $

Answer 1

Hint: To solve this question, we will take the given equation and simplify it using the double angle formulas of sine. After this using only two components of Law of sine that is $\frac{a}{\sin A}=\frac{b}{\sin B}$we will derive a relation and substitute in the equation. Then we will substitute the projection formula in the equation and simplify it. After this using formula of area of triangle we will derive another relation and substitute it and simplify it and get the required value.
Formula used:
The double angle formula of sine is:
$\sin 2A=2\sin A\cos A$
Projection formula:
$a\cos B+b\cos A=c$
Area of triangle:
$\Delta =\frac{1}{2}ac\sin B$
Complete step-by-step solution:
We are given area of a triangle $ABC$as $\Delta $ and we have to calculate the value of ${{a}^{2}}\sin 2B+{{b}^{2}}\sin 2A$. We will take the equation and use the double angle formula of sine.
$\begin{align}
  & ={{a}^{2}}\left( 2\sin B\cos B \right)+{{b}^{2}}\left( 2\sin A\cos A \right) \\
 & =2{{a}^{2}}\sin B\cos B+2{{b}^{2}}\sin A\cos A.....(i) \\
\end{align}$
We will now use the Law of sine $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$by taking only two component of this relationship that is $\frac{a}{\sin A}=\frac{b}{\sin B}$.
$a\sin B=b\sin A$
We will now use the derived relation $a\sin B=b\sin A$in equation (i).
$\begin{align}
  & =2{{a}^{2}}\sin B\cos B+2b\times b\sin A\times \cos A \\
 & =2{{a}^{2}}\sin B\cos B+2b\times a\sin B\times \cos A \\
 & =2a\sin B(a\cos B+b\cos A) \\
\end{align}$
Now substituting the projection formulas in the equation we will get,
$\begin{align}
  & =2a\sin B\times c \\
 & =2ac\sin B......(ii) \\
\end{align}$
We will now use the formula of area of the triangle with two sides and an angle and derive the value of $ac\sin B$.
$\begin{align}
  & \Delta =\frac{1}{2}ac\sin B \\
 & ac\sin B=2\Delta
\end{align}$
Now we will substitute the above derived values in equation (ii).
$\begin{align}
  & =2ac\sin B \\
 & =2\times 2\Delta \\
 & =4\Delta \\
\end{align}$
The value of ${{a}^{2}}\sin 2B+{{b}^{2}}\sin 2A$ is $4\Delta $ when the area of the triangle is $\Delta $.Hence the correct option is (C).
Note:
There are three formulas for the area of the triangle with different sides and angle of the triangle. In the above question we chose the formula $\Delta =\frac{1}{2}ac\sin B$ because the angle and sides are similar to the other simplified equation. Hence we should choose accordingly.