Question

In a acute angle triangle ABC, if $sin(A+B-C)=\frac{1}{2}$ and $cos(B+C-A)=\frac{1}{\sqrt{2}}$.

Find $\angle A,\angle B$ and $\angle C$.

Answer 1

Given,

$A+B+C={180}^0⟶\left(1\right)$

$\sin (A+B-C)=\frac{1}{2}⟶\left(2\right)$

$\cos (B+C-A)=\frac{1}{\sqrt{2}}⟶\left(3\right)$

Now from $\left(1\right)$

$A+B=\Pi -C⟶\left(4\right)$

$\therefore$ Using $\left(4\right)$ we can rewrite $\left(2\right)$ as

$\sin (\Pi -2C)=\frac{1}{2}$

$\Rightarrow \sin 2C=\frac{1}{2}(\because \sin (\Pi -\theta )=\sin \theta )$

$\therefore$ $2C={30}^0$ or ${150}^0$ ($\because$ ${\sin 30}^0=1/2$

$\Rightarrow C={15}^0$ or ${75}^0$ ${\sin 150}^0=1/2$)

Again from $\left(1\right)$

$B+C=\Pi -A$

$\therefore$ $\cos (B+C-A)=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos (\Pi -2A)=\frac{1}{\sqrt{2}}$

$\Rightarrow -\cos 2A=\frac{1}{\sqrt{2}}$ ($\because$ $\cos (\Pi -\theta )=-\cos \theta$)

$\Rightarrow \cos 2A=\frac{-1}{\sqrt{2}}$

$\therefore$ $2A={135}^0$ ($\because$ ${\cos 135}^0=\frac{-1}{\sqrt{2}}$)

$\Rightarrow A={67.5}^0$

Case I : For $C={15}^0$ Case II : For $C={75}^0$

$A={67.5}^0$ $A={67.5}^0$

$\therefore$ $B={180}^0-({15}^0+{67.5}^0)$ $B={180}^0-({67.5}^0+{75}^0)$

$={180}^0-82.5={97.5}^0$ $={37.5}^0$

Since $\Delta ABC$ is acute angled, thereby $n$ none of the angles can be more than ${90}^0$.

Hence $\angle A={67.5}^0$, $\angle C={75}^0$, $\angle B={37.5}^0$