A) \[15{}^\circ \]
B) \[30{}^\circ \]
C) \[60{}^\circ \]
D) \[75{}^\circ \]
Correct Answer: C
Solution :
[c] We have \[AC=\sec \theta ,AG=8\] \[\therefore CG=8-\sec \theta \] (C being the peg). But \[u=CG\sin \theta =(8-sec\theta )sin\theta \] \[u=8\sin \theta -\tan \theta \] \[\frac{du}{d\theta }=8\cos \theta -{{\sec }^{2}}\theta ,\] \[\frac{{{d}^{2}}u}{d{{\theta }^{2}}}=-8\sin \theta -2{{\sec }^{2}}\theta \tan \theta \] \[\frac{du}{d\theta }=0,\] when \[{{\cos }^{3}}\theta =\frac{1}{8},\cos \theta =\frac{1}{2},\] \[\frac{{{d}^{2}}u}{d{{\theta }^{2}}}>0(at\theta =60{}^\circ ),\therefore \theta =60{}^\circ \]You need to login to perform this action.
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