| If \[\tan \theta =\frac{3}{4}\]and \[0<\theta <\frac{\pi }{2}\]and \[25x{{\sin }^{2}}\theta \cos \theta ={{\tan }^{2}}\theta ,\]then the value of \[x\]is [SSC (CGL) 2014] |
A) \[\frac{7}{64}\]
B) \[\frac{9}{64}\]
C) \[\frac{3}{64}\]
D) \[\frac{5}{64}\]
Correct Answer: D
Solution :
| Given, \[\tan \theta =\frac{3}{4}=\frac{p}{b}\] |
| Then,\[h=\sqrt{{{p}^{2}}+{{b}^{2}}}=\sqrt{9+16}=\sqrt{25}=5\] |
| \[\therefore \]\[\sin \theta =\frac{p}{h}=\frac{3}{5}\] |
| \[\Rightarrow \]\[\cos \theta =\frac{b}{h}=\frac{4}{5}\] |
| Now, \[25x{{\sin }^{2}}\theta \cos \theta ={{\tan }^{2}}\theta \] |
| \[\Rightarrow \]\[25\cdot x\cdot {{\left( \frac{3}{5} \right)}^{2}}\cdot \frac{4}{5}={{\left( \frac{3}{4} \right)}^{2}}\] |
| \[\Rightarrow \]\[25\cdot x\cdot \frac{9}{25}\cdot \frac{4}{5}=\frac{9}{16}\] |
| \[\therefore \] \[x=\frac{5}{64}\] |
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