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Step-by-step Answer
PAGE: 90SET: ExercisesPROBLEM: 71
Please look in your text book for this problem Statement

The base of a right triangle is 10 mts.

The radius of a circle is 10 mts.

(a)

The area of triangle =\frac{1}{2}bh.

The area of sector =\frac{1}{2}r^{2}\theta.

The area of shaded region = area of triangle - area of sector.

The area of shaded region f(\theta )=\frac{1}{2}bh-\frac{1}{2}r^{2}\theta.

In the triangle,

\tan \theta =\frac{\textrm{height}}{\textrm{base}}=\frac{h}{10}

h=10\tan \theta.

Substitute the values in the function.

\\f(\theta )=\frac{1}{2}bh-\frac{1}{2}r^{2}\theta\\\\ =\frac{1}{2}\left ( 10 \right )\left ( 10\tan \theta \right )-\frac{1}{2}10^{2}\theta\\\\ =50\tan \theta -50\theta

Domain of \\f(\theta ) =50\tan \theta -50\theta is \left ( 0,\ \frac{\pi}{2}\right ).

(b)Graph the function \\f(\theta ) =50\tan \theta -50\theta.

Observe the graph:

Complete the table.

\theta 0.3 0.6 0.9 1.2 1.5
f(\theta ) 0.4668 4.2068 18.0079 68.6075 630.0709

(c) Find \lim_{\theta \rightarrow \frac{\pi}{2}^-}f(\theta ).

\lim_{\theta \rightarrow \frac{\pi}{2}^-} 50\tan \theta -50\theta

As x tends to  \frac{\pi}{2} from left hand side, the function approaches to \infty.

\lim_{\theta \rightarrow \frac{\pi}{2}^-}f(\theta )=\lim_{\theta \rightarrow \frac{\pi}{2}^-} 50\tan \theta -50\theta=\infty.

(a) Domain of \\f(\theta ) =50\tan \theta -50\theta , Domain: \left ( 0,\ \frac{\pi}{2}\right ).

(b)

\theta 0.3 0.6 0.9 1.2 1.5
f(\theta ) 0.4668 4.2068 18.0079 68.6075 630.0709

(c) \lim_{\theta \rightarrow \frac{\pi}{2}^-}f(\theta )=\lim_{\theta \rightarrow \frac{\pi}{2}^-} 50\tan \theta -50\theta=\infty.

 



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