Question

If $\Delta ={{a}^{2}}-{{(b-c)}^{2}}$, where $\Delta$ is the area of the triangle $ABC$, then $\tan A$ is equal to
A. $\dfrac{15}{16}$
B. $\dfrac{8}{15}$
C. $\dfrac{8}{17}$
D. $\dfrac{1}{2}$

Answer 1

Hint: To solve this question, we will simplify the given equation and derive an equation in terms of the semi perimeter of the triangle $s$ and consider it as the first equation. We will then use the half angle formula of tan and simplify it using Heron’s formula and deduce an equation for $\tan \dfrac{A}{2}$ and consider it as the second equation. Then we will compare both the equations and derive the numerical value of $\tan \dfrac{A}{2}$. We will then substitute this value of $\tan \dfrac{A}{2}$ in the formula of $\tan A$ and find its value.

Formula Used: ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$,
Semi perimeter of the triangle is:
$\begin{align}
  & s=\dfrac{a+b+c}{2} \\
 & 2s=a+b+c \\
\end{align}$
The formulas of tan are:
$\tan \dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$
$\tan A=\dfrac{2\tan \dfrac{A}{2}}{1-{{\tan }^{2}}\dfrac{A}{2}}$
Heron’s Formula:
$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$

Complete step by step solution: We are given area of triangle$ABC$ , $\Delta ={{a}^{2}}-{{(b-c)}^{2}}$ where $\Delta $is the area of the triangle and we have to calculate the value of $\tan A$.
We will take the given area of the triangle and simplify it using formula of ${{a}^{2}}-{{b}^{2}}$.
$\begin{align}
  & \Delta ={{a}^{2}}-{{(b-c)}^{2}} \\
 & =(a-(b-c)(a+(b-c)) \\
 & =(a-b+c)(a+b-c)
\end{align}$
Now we can write the equation as,
$\Delta =(a+b+c-2b)(a+b+c-2c)$
We will now substitute the formula of the semi perimeter of the triangle $2s=a+b+c$.
$\begin{align}
  & \Delta =(2s-2b)(2s-2c) \\
 & \Delta =4(s-b)(s-c) \\
 & \dfrac{1}{4}=\dfrac{(s-b)(s-c)}{\Delta }....(i)
\end{align}$
Now we will use the formula of $\tan \dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$ and write it in the form of equation we derived from the area of the triangle that was given.
$\tan \dfrac{A}{2}\sqrt{s(s-a)}=\sqrt{(s-b)(s-c)}$
We will now multiply this equation on both sides by $\sqrt{(s-b)(s-c)}$.
$\begin{align}
  & \tan \dfrac{A}{2}\sqrt{s(s-a)}\times \sqrt{(s-b)(s-c)}=\sqrt{(s-b)(s-c)}\times \sqrt{(s-b)(s-c)} \\
 & \tan \dfrac{A}{2}\sqrt{s(s-a)(s-b)(s-c)}=(s-b)(s-c)
\end{align}$
We know that $\sqrt{s(s-a)(s-b)(s-c)}$ is the Heron’s formula to calculate the area of the triangle hence,
\[\begin{align}
  & \tan \dfrac{A}{2}.\Delta =(s-b)(s-c) \\
 & \tan \dfrac{A}{2}=\dfrac{(s-b)(s-c)}{\Delta }.....(ii) \\
\end{align}\]
Comparing both the equation (i) and (ii) we will get,
$\tan \dfrac{A}{2}=\dfrac{1}{4}$
We will now use the formula $\tan A=\dfrac{2\tan \dfrac{A}{2}}{1-{{\tan }^{2}}\dfrac{A}{2}}$ and substitute the value we derived$\tan \dfrac{A}{2}=\dfrac{1}{4}$.
$\begin{align}
  & \tan A=\dfrac{2\times \dfrac{1}{4}}{1-{{\left( \dfrac{1}{4} \right)}^{2}}} \\
 & =\dfrac{\dfrac{1}{2}}{\dfrac{15}{16}} \\
 & =\dfrac{8}{15}
\end{align}$

The value of $\tan A$ is $\tan A=\dfrac{8}{15}$ when the area of the triangle $ABC$ is $\Delta ={{a}^{2}}-{{(b-c)}^{2}}$. Hence the correct option is (B).

Note: We should notice that a lot of formulas are used here from the semi perimeter of the triangle, Heron’s Formula, Expansion formula and all the other trigonometric formulas so we must remember all these to solve these kinds of questions.