question_answer

In \[\Delta ABC,\angle B={{90}^{o}}\]  and b + a = 4, the area of the triangle is the maximum, when \[\angle C\] is

A)  \[\frac{\pi }{4}\]                                             

B)  \[\frac{\pi }{6}\]

C)  \[\frac{\pi }{3}\]                                             

D)  None of these

Correct Answer: C

Solution :

In \[\Delta ACB,\] \[b\cos \theta =a\]                                         ?(i) Given, \[b\cos \theta +b=4\]\[(\because b+a=4\text{given})\] \[b=\frac{4}{1+\cos \theta }\] From Eq. (i), \[a=\frac{4\cos \theta }{1+\cos \theta }\] \[\therefore \]Area of triangle, \[\Delta =\frac{1}{2}ba\sin \theta \] \[=\frac{1}{2}.\frac{4}{1+\cos \theta }.\frac{4\cos \theta }{1=\cos \theta }\sin \theta \] \[=\frac{4\sin 2\theta }{{{(1+\cos \theta )}^{2}}}\] \[\therefore \]\[\frac{d\Delta }{d\theta }=4\frac{\left[ \begin{align}   & 2\cos 2\theta {{(1+\cos \theta )}^{2}} \\  & +\sin 2\theta .2(1+\cos \theta )\sin \theta  \\ \end{align} \right]}{{{(1+\cos \theta )}^{4}}}\] \[=\frac{8[\cos 2\theta (1+\cos \theta )+\sin 2\theta \sin \theta ]}{{{(1+\cos \theta )}^{3}}}\] For maxima or minima, Put\[\frac{d\Delta }{d\theta }=0\] \[\Rightarrow \]\[\cos 2\theta (1+\cos \theta )+\sin 2\theta \sin \theta =0\] \[\Rightarrow \]\[\cos 2\theta +(\cos 2\theta \cos \theta +\sin 2\theta \sin \theta )=0\] \[\Rightarrow \]\[\cos 2\theta +\cos \theta =0\] \[\therefore \]\[\cos 2\theta =-\cos \theta =\cos (\pi -\theta )\] \[\Rightarrow \]\[\theta =\frac{\pi }{3}\] \[\therefore \]\[\Delta \] is maximum when \[\theta =\frac{\pi }{3}.\]