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

We have ,
tan( A + B - C) = 1 and sec (B + C - A) = 2

$\Rightarrow$ tan( A + B - C) = $tan{45}^{\circ }$ and sec (B + C - A) = $sec{60}^{\circ }$

$\Rightarrow$ A + B - C =${45}^{\circ }$ and B + C - A = ${60}^{\circ }$
$\Rightarrow$ ( A + B - C) + (B + C - A) = ${45}^{\circ }+{60}^{\circ }$
$\Rightarrow$ 2B = 105${}^{\circ }$

$\Rightarrow B=52\frac{1^{\circ }}{2}$
Putting $B=52\frac{1^{\circ }}{2}$ in B + C - A = 60${}^{\circ }$ , we get

$52\frac{1^{\circ }}{2}+C-A={60}^{\circ }\Rightarrow C-A=7\frac{1^{\circ }}{2}$ ...(i)
Also , in $\Delta ABC$ , we have
A + B + C = ${180}^{\circ }$

$\Rightarrow A+52\frac{1^{\circ }}{2}+C={180}^{\circ }$ $[\because B=52\frac{1^{\circ }}{2}]$

$\Rightarrow C+A=127\frac{1^{\circ }}{2}$ ...(ii)
Adding and subtracting (i) and (ii) , we get ,
2C = 135${}^{\circ }$ and 2A = 120${}^{\circ }\Rightarrow C=67\frac{1^{\circ }}{2}$ and A = 60${}^{\circ }$

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