Question

The area of the triangle ABC with altitude AD $\angle BAC={{45}^{\circ }},DB=3,BC=2$ units is \[\]
A.6\[\]
B.15\[\]
C.$\dfrac{15}{4}$\[\]
D.12\[\]

Answer 1

\[\] Hint: We take $\angle BAC=\angle BAD\angle CAD=x+y,AD=l$. We take tangent trigonometric function on both sides and use the tangent of sum of two angles formula $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. We put the data given in the equation and find the quadratic equation in $l$. We solve the equation to get $l$ and find the area as $\dfrac{1}{2}\times BC\times l$.\[\]

Complete step by step answer:
We know that the tangent of sum two angles say $A,B$ is given by the formula
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]

We have drawn as described in the question where AD is an altitude or perpendicular on BC in the triangle ABC. We are given that $\angle BAC={{45}^{\circ }}$, BD=3 units, CD=2 units. So we have $BC=BD+CD=3+2=5$ units. We know that the formula for the area of the triangle is given as half the product of base and perpendicular. So we have area $\Delta $ of the triangle ABC as
\[\Delta =\dfrac{1}{2}\times BC\times AD\]

Now we need to only find the length of AD. We know that in a right angled triangle the tangent of any angle is the ratio of lengths of opposite side and the adjacent side excluding hypotenuse . Let us assume $\angle BAD=x,\angle CAD=y$ and the length of AD as $l$. We take the tangent of angle $\angle BAD$ in the right angled triangle BAD and get ,
\[\begin{align}
  & \tan \left( \angle BAD \right)=\dfrac{BD}{AD} \\
 & \Rightarrow \tan x=\dfrac{3}{l} \\
\end{align}\]
We take the tangent of angle $\angle CAD$ in the right angled triangle CAD and get ,
\[\begin{align}
  & \tan \left( \angle CAD \right)=\dfrac{CD}{AD} \\
 & \Rightarrow \tan y=\dfrac{2}{l} \\
\end{align}\]
We have from the triangle ABC, $\angle BAC=\angle BAD+\angle CAD$. Let us take tangent at both side and get,
\[\begin{align}
  & \angle BAC=\angle BAD+\angle CAD \\
 & \Rightarrow \tan \left( \angle BAC \right)=\tan \left( \angle BAD+\angle CAD \right) \\
 & \Rightarrow \tan {{45}^{\circ }}=\tan \left( x+y \right) \\
\end{align}\]
We know that $\tan {{45}^{\circ }}=1$. We use the tangent of sum of two angles formula and get
\[\begin{align}
  & \Rightarrow 1=\dfrac{\tan x+\tan y}{1-\tan x\tan y} \\
 & \Rightarrow \dfrac{\dfrac{3}{l}+\dfrac{2}{l}}{1-\dfrac{3}{l}\cdot \dfrac{2}{l}}=1 \\
 & \Rightarrow \dfrac{5}{l}=1-\dfrac{6}{{{l}^{2}}} \\
 & \Rightarrow {{l}^{2}}-5l-6=0 \\
\end{align}\]
We split the middle term and solve the above quadratic equation.
\[\begin{align}
  & \Rightarrow {{l}^{2}}-5l-6=0 \\
 & \Rightarrow {{l}^{2}}-6l+l-6=0 \\
 & \Rightarrow l\left( l-6 \right)+1\left( l-6 \right)=0 \\
 & \Rightarrow \left( l-6 \right)\left( l+1 \right)=0 \\
 & \Rightarrow l=6,l=-1 \\
\end{align}\]
We reject the negative value for distance and we find $l=6=AD$. So the area of the triangle in square units is

\[\Delta =\dfrac{1}{2}\times BC\times AD=\dfrac{1}{2}\times 5\times 6=15\]
So the correct option is B. \[\]

Note:
We note that the formula for tangent of sum of two angles is valid when $A\ne {{90}^{\circ }},B\ne {{90}^{\circ }},A+B<{{180}^{\circ }}$. If $A+B>{{180}^{\circ }}$ we have $\tan \left( A+B \right)=\pi -\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ . We find the acute angle between two lines with slope ${{m}_{1}}=\tan A$ and ${{m}_{2}}=\tan B$ using ${{\tan }^{-1}}\left( \dfrac{{{m}_{1}}+{{m}_{2}}}{1-{{m}_{1}}{{m}_{2}}} \right)$.