Question

If in a triangle $ABC,a=6,b=3$ and $\cos (A-B)=\frac{4}{5}$ then find its area.

• A. $9\sqrt{2}$ sq. units
• B. $9$ sq. units
• C. $\frac{9}{\sqrt{2}}$ sq. units
• D. $18$ sq. units

Answer 1

Answer: (C)

$\frac{9}{\sqrt{2}}$ sq. units

Given, $a=6,b=3$ and $\cos (A-B)=\frac{4}{5}$

We have the formula for area of triangle, $Area=\frac{1}{2}ab\sin C$

We know the below formulas,

$\tan \left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\left(\cot C\right)$ ------ (1)

$\tan \left(\frac{A-B}{2}\right)=\sqrt{\frac{1-\cos (A-B)}{1+\cos (A-B)}}$ ------ (2)

Substitute for $a,b$ in (1)

$\tan \left(\frac{A-B}{2}\right)=\frac{6-3}{6+3}\left(\cot C\right)=\frac{1}{3}\left(\cot C\right)$ ---- (3)

Substitute for $\cos (A-B)$ in (2)

$\tan \left(\frac{A-B}{2}\right)=\sqrt{\frac{1-\frac{4}{5}}{1+\frac{4}{5}}}=\sqrt{\frac{1}{9}}=\frac{1}{3}$ ---- (4)

From (3) and (4)

$\frac{1}{3}\left(\cot C\right)=\frac{1}{3}$

$\cot C=1$

$⟹\cot C={45}^{\circ }$

$\therefore Area=\frac{1}{2}ab\sin C$

$=\frac{1}{2}\times 6\times 3\times \sin {45}^{\circ }$

$=\frac{1}{2}\times 6\times 3\times \frac{1}{\sqrt{2}}$

$\therefore Area=\frac{9}{\sqrt{2}}Sq.units$