Question

If in a $\Delta ABC$ ,a = 6 , b = 3 and cos ( A – B ) = $\dfrac{4}{5}$ then find its area ?

Answer 1

Hint: With the given details we can apply it in the formula \[\tan \left( {\dfrac{{A - B}}{2}} \right) = \sqrt {\dfrac{{1 - \cos \left( {A - B} \right)}}{{1 + \cos \left( {A - B} \right)}}} \]and we know that $\tan \left( {\dfrac{{A - B}}{2}} \right) = \dfrac{{a - b}}{{a + b}}\cot \dfrac{C}{2}$using which we can find the value of C and since its 90 degree we can find the area using the formula $\dfrac{1}{2}*base*height$.

Complete step-by-step answer:
We are given that a = 6 cm and b = 3 cm and cos ( A – B ) = $\dfrac{4}{5}$
We know that \[\tan \left( {\dfrac{{A - B}}{2}} \right) = \sqrt {\dfrac{{1 - \cos \left( {A - B} \right)}}{{1 + \cos \left( {A - B} \right)}}} \]
Now applying the given value we get,
$
\Rightarrow \tan \left( {\dfrac{{A - B}}{2}} \right) = \sqrt {\dfrac{{1 - \dfrac{4}{5}}}{{1 + \dfrac{4}{5}}}} \\
 \Rightarrow \tan \left( {\dfrac{{A - B}}{2}} \right) = \sqrt {\dfrac{{\dfrac{{5 - 4}}{5}}}{{\dfrac{{5 + 4}}{5}}}} \\
   \Rightarrow \tan \left( {\dfrac{{A - B}}{2}} \right) = \sqrt {\dfrac{1}{9}} = \dfrac{1}{3} \\
$
Using this we can substitute in the formula
$ \Rightarrow \tan \left( {\dfrac{{A - B}}{2}} \right) = \dfrac{{a - b}}{{a + b}}\cot \dfrac{C}{2}$
Substituting all the known values,
$
   \Rightarrow \dfrac{1}{3} = \dfrac{{6 - 3}}{{6 + 3}}\cot \dfrac{C}{2} \\
   \Rightarrow \dfrac{1}{3} = \dfrac{3}{9}\cot \dfrac{C}{2} \\
   \Rightarrow \dfrac{1}{3} = \dfrac{1}{3}\cot \dfrac{C}{2} \\
   \Rightarrow \cot \dfrac{C}{2} = 1 \\
   \Rightarrow \dfrac{C}{2} = {\cot ^{ - 1}}(1) = \dfrac{\pi }{4} \\
   \Rightarrow C = \dfrac{{2\pi }}{4} = \dfrac{\pi }{2} \\
$
From this we get that the angle C is a right angle
Therefore the triangle ABC is a right angle triangle , right angled at B

Therefore the area of the triangle = $\dfrac{1}{2}*base*height$ sq . units
=$\dfrac{1}{2}*a*b$
 =$\dfrac{1}{2}*6*3 = \dfrac{1}{2}*18 = 9sq.units$
Therefore the area of the triangle is 9 sq.units


Note: Here we get to know that it’s a right triangle hence we use the side AC as its height but if it is not an right triangle the altitude may vary.