Question

In a $\Delta ABC$, if $A=ta{n}^{-1}2$ and $B=ta{n}^{-1}3$ , then $C=$

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{3\pi }{4}$

Answer 1

Answer: (B)

We have, $A=ta{n}^{-1}2\Rightarrow tanA=2$
and $B=ta{n}^{-1}3\Rightarrow tanB=3$
since $A,B,C$ are angles of a triangle, $A+B+C=\pi$
$\Rightarrow C=\pi -\left(A+B\right)$
Now, $A+B=ta{n}^{-1}2+ta{n}^{-1}3=\pi +ta{n}^{-1}\left[\frac{2+3}{1-2.3}\right]$
$\left[\therefore ta{n}^{-1}x+ta{n}^{-1}y=\pi +ta{n}^{-1}\left[\frac{x+y}{1-xy}\right]<br/>forx>0,y>0andxy>1\right]$
$=\pi +ta{n}^{-1}\left(-1\right)=\pi -ta{n}^{-1}1=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$
$\therefore From\left(1\right),C=\pi -\frac{3\pi }{4}=\frac{\pi }{4}$