Question

In $\Delta ABC$, if $a=5;b=4,A=\frac{\pi }{2}+B$ then $\tan C=$

• A. $\frac{9}{40}$
• B. $4$
• C. $2.5$
• D. $1.45$

Answer 1

Answer: (A)

$\frac{9}{40}$

Given, $A-B={90}^o,a=5,b=4;$

From Napier's Formula

$\Rightarrow \tan \left(\frac{{90}^o}{2}\right)=\frac{5-4}{5+4}\cot \frac{C}{2}\Rightarrow \cot \frac{C}{2}=9\Rightarrow \tan \frac{C}{2}=\frac{1}{9}$

we know that $\tan C=\frac{2\tan \left(C/2\right)}{1-{\tan }^2\left(C/2\right)}=\frac{9}{40}$

Therefore , Answer is $A$