Question

In an acute angled triangle ABC, if $\tan (A+B-C)=1$ and, $\sec (B+C-A)=2$, find the value of A, B and C.

Answer 1

$\tan (A+B-C)=1=\tan {45}^o$

$A+B-C={45}^o$ .........(1)

$\sec (B+C-A)=2=\sec {60}^o$

$B+C-A={60}^o$ ........(2)

Solving equations 1 and 2, we get,

$B=52{\frac{1}{2}}^o$

$C-A=7{\frac{1}{2}}^o$ ........(3)

In $△ABC$,

$A+B+C={180}^o$

$C+A=127{\frac{1}{2}}^o$ ........(4)

Solving equations 3 and 4, we get,

$A={60}^o$ and $C=67{\frac{1}{2}}^o$

Hence, $A={60}^o,B=52{\frac{1}{2}}^o$ and $C=67{\frac{1}{2}}^o$