Question

In each of $\alpha$ $\beta$ and $\gamma$ $\text{is a positive acute angle such that}$
$\sin (\alpha +\beta -\gamma )=\frac{1}{2},\cos (\beta +\gamma -\alpha )=\frac{1}{2}\text{and}\tan (\gamma +\alpha -\beta )=1$
Find the values of $\alpha$, $\beta$, & $\gamma$.

Answer 1

We have ,

$\sin (\alpha +\beta -\gamma )=\frac{1}{2},\cos (\beta +\gamma -\alpha )=\frac{1}{2}\text{and}\tan (\gamma +\alpha -\beta )=1$

Lets take, $\sin (\alpha +\beta -\gamma )=\frac{1}{2}$

$\Rightarrow \sin (\alpha +\beta -\gamma )=\sin {30}^{\circ }$

$\Rightarrow \alpha +\beta -\gamma ={30}^{\circ }$---(1)

Lets take, $\cos (\beta +\gamma -\alpha )=\frac{1}{2}$

$\cos (\beta +\gamma -\alpha )=\cos {60}^{\circ }$

$\Rightarrow \beta +\gamma -\alpha ={60}^{\circ }$---(2)

Lets take, $\tan (\gamma +\alpha -\beta )=1$

$\tan (\gamma +\alpha -\beta )=\tan {45}^{\circ }$

$\Rightarrow \gamma +\alpha -\beta ={45}^{\circ }$ ----(3)

Now, lets add (1) and (2)

$\alpha +\beta -\gamma +\beta +\gamma -\alpha ={30}^{\circ }+{60}^{\circ }$

$\Rightarrow 2\beta ={90}^{\circ }$

$\Rightarrow \beta =\frac{{90}^{\circ }}{2}$

$\Rightarrow \beta ={45}^{\circ }$ ---(4)

Lets add (2) and (3)

$\beta +\gamma -\alpha +\gamma +\alpha -\beta ={60}^{\circ }+{45}^{\circ }$

$\Rightarrow 2\gamma ={105}^{\circ }$

$\Rightarrow \gamma =\frac{{105}^{\circ }}{2}$

$\Rightarrow \gamma =52{\frac{1}{2}}^{\circ }$ --(5)

Now, substitute $\beta ={45}^{\circ },\gamma =52{\frac{1}{2}}^{\circ }$ in $\alpha +\beta -\gamma ={30}^{\circ }$

$\Rightarrow \alpha +\beta -\gamma ={30}^{\circ }$

$\Rightarrow \alpha +{45}^{\circ }-52{\frac{1}{2}}^{\circ }={30}^{\circ }$

$\Rightarrow \alpha -7{\frac{1}{2}}^{\circ }={30}^{\circ }$

$\Rightarrow \alpha ={30}^{\circ }+7{\frac{1}{2}}^{\circ }$

$\Rightarrow \alpha =37{\frac{1}{2}}^{\circ }$

Hence , $\alpha =37\frac{1^{\circ }}{2},\beta ={45}^{\circ }and\gamma =52\frac{1^{\circ }}{2}$