A) \[1.2\,V\]
B) \[2.4\,V\]
C) \[3.6\,V\]
D) \[4.8\,V\]
Correct Answer: A
Solution :
\[\therefore \,\,h=\frac{a\cos \theta }{3}\Rightarrow \cos \theta =\frac{3h}{a}\] \[k=\frac{b\sin \theta }{3}\Rightarrow \,\sin \theta =\frac{3k}{b}\] \[\therefore \,\,\frac{{{x}^{2}}}{{{a}^{2}}}\,+\frac{{{y}^{2}}}{{{b}^{2}}}=\frac{1}{9}\] \[\int\limits_{0}^{\pi /6}{\,\frac{\cos x-\sin x}{1+\sin 2x}dx=2-\sqrt{3}}\]
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