A) \[\frac{45}{16}\]
B) \[\frac{35}{18}\]
C) \[\frac{32}{16}\]
D) \[\frac{47}{16}\]
Correct Answer: D
Solution :
Given, \[4\,\sin \theta =3\] |
\[\Rightarrow \,\,\,\sin \theta =\frac{3}{4}=\frac{Opposite\,side}{Hypotenuse}\] |
Let \[AB=3k\]and \[AC=4k\] |
Apply Pythagoras theorem, |
\[A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}\] |
\[\Rightarrow \,\,\,\,\,\,\,{{\left( 3k \right)}^{2}}+B{{C}^{2}}={{\left( 4k \right)}^{2}}\] |
\[\Rightarrow \,\,\,\,\,BC=\left( \sqrt{16-9} \right)k\] |
\[=\sqrt{7}\,\,\,k\] |
\[\Rightarrow \,\,\,\cos \theta =\frac{\sqrt{7}k}{4k}=\frac{\sqrt{7}}{4}\] |
\[\Rightarrow \,\,\,\,\,\,4{{\sin }^{2}}\theta -3{{\cos }^{2}}\theta +2=4\times \frac{9}{16}-3\times \frac{7}{16}+2\] |
\[=\frac{36-21+32}{16}=\frac{47}{16}\] |
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