Question

If $\sin \left({\theta }_1-{\theta }_2\right)=\frac{1}{2}$ and $\cos \left({\theta }_1+{\theta }_2\right)=\frac{1}{2},0°<{\theta }_1+{\theta }_2<90°,{\theta }_1>{\theta }_2$, then find ${\theta }_1$and ${\theta }_2$.

Answer 1

Step-1 Forming equation in ${\theta }_1$ and ${\theta }_2$

$\begin{array}{rcl}\sin \left({\theta }_1-{\theta }_2\right)& =& \frac{1}{2}\\ & \Rightarrow & \sin \left({\theta }_1-{\theta }_2\right)=\sin 30°\\ \Rightarrow {\theta }_1-{\theta }_2& =& 30°...\left(1\right)\end{array}\left[\because \sin 30°=\frac{1}{2}\right]$

Similarly,

$\begin{array}{rcl}\cos \left({\theta }_1+{\theta }_2\right)& =& \frac{1}{2}\\ & \Rightarrow & \cos \left({\theta }_1+{\theta }_2\right)=\cos 60°\\ \Rightarrow {\theta }_1+{\theta }_2& =& 60°...\left(2\right)\end{array}\left[\because \cos 60°=\frac{1}{2}\right]$

Step-2 Solution of equations:

Now,

${\theta }_1-{\theta }_2=30°...\left(1\right)$

${\theta }_1+{\theta }_2=60°...\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow 2{\theta }_1=90°$

$\Rightarrow {\theta }_1=45°$

Putting ${\theta }_1=45°$in $\left(2\right)$

$45°+{\theta }_2=60°$

$\Rightarrow {\theta }_2=15°$

Hence, the values of ${\theta }_1$ and ${\theta }_2$ is $45°$ and $15°$ respectively.