Question

If the area AMBECL is 1/nth of the field, then $sin\theta +(\pi -\theta )cos\theta$ is equal to

• A. $n\pi$
• B. $\frac{n-1}{n}\pi$
• C. $(n-1)\pi$
• D. $(n+1)\pi$

Answer 1

The correct option is B $\frac{n-1}{n}\pi$

$\angle AOB=\theta \Rightarrow \angle OAB+\angle OBA=\pi -\theta \Rightarrow \angle OAB=\frac{\pi }{2}-\frac{\theta }{2}$
Gives $AB=2Rcos(\frac{\pi }{2}-\frac{\theta }{2})$ ...(1)
Area of segment AMB = Area of sector AMB - Area of $\Delta AOB$
$AB=2Rcos(\frac{\pi }{2}-\frac{\theta }{2})=2Rsin\left(\theta /2\right)=\frac{1}{2}{R}^2\theta -\frac{1}{2}AB\times Rsin(\frac{\theta }{2}-\frac{\pi }{2})$
$=\frac{1}{2}{R}^2\theta -\frac{1}{2}\times 2R\left(\frac{\theta }{2}\right)\times Rcos\left(\frac{\theta }{2}\right)=\frac{1}{2}{R}^2(\theta -sin\theta )$ ...(2)
Now, Area of AMBECL =2( Area of AMBEOA)
$\Rightarrow \frac{1}{n}\pi R^2=2$ Area of sector ABE + Area of segment AMB
$=\frac{1}{2}{R}^2\theta -\frac{1}{2}\times 2R\left(\frac{\theta }{2}\right)\times Rcos\left(\frac{\theta }{2}\right)=\frac{1}{2}{R}^2(\theta -sin\theta )\Rightarrow \frac{1}{n}\pi R^2=2\left[\frac{1}{2}\right(AB{)}^2(\frac{\pi }{2}-\frac{\theta }{2})+\frac{1}{2}{R}^2(\theta -sin\theta )]=2[\frac{1}{2}\times 4R^{2}si{n}^2\frac{\theta }{2}(\frac{\pi }{2}-\frac{\theta }{2})+\frac{1}{2}{R}^2(\theta -sin\theta \left)\right]=R^2\left[\right(1-cos\theta \left)\right(\pi -\theta )+\theta -sin\theta ]=R^2[\pi -(\pi -\theta )cos\theta -sin\theta$