Question

The area of a triangle $ABC$ is equal to $\left(a^2+b^2-c^2\right)$, where $a,b$ and $c$ are the sides of the triangle. The value of $\tan C$ equals

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (B)

$\cos C=\frac{a^2+b^2-c^2}{2ab}$
Now, $\Delta =a^2+b^2-c^2$
Hence, $\cos C=\frac{\Delta }{2ab}$ ...(1)
Also, $\Delta =\frac{1}{2}ab\sin C$
$\Rightarrow \frac{2\Delta }{\sin C}=ab$
$\Rightarrow \sin C=\frac{2\Delta }{ab}$ ...(2)
From (1) and (2), we get
$\tan C=\frac{2\Delta }{ab}\cdot \frac{2ab}{\Delta }=4$