A) \[{{\tan }^{-1}}\left( \frac{{{t}_{1}}}{{{t}_{2}}} \right)\]
B) \[{{\sin }^{-1}}\left( \frac{{{t}_{1}}}{{{t}_{2}}} \right)\]
C) \[{{\sin }^{-1}}\left( \frac{10{{t}_{1}}}{{{t}_{2}}} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{10{{t}_{1}}}{{{t}_{2}}} \right)\]
Correct Answer: C
Solution :
From the formula \[\sin C=\frac{1}{_{1}{{\mu }_{2}}}\]\[\Rightarrow \sin C{{=}_{2}}{{\mu }_{1}}\] \[=\frac{{{u}_{1}}}{{{u}_{2}}}=\frac{{{v}_{2}}}{{{v}_{1}}}\]\[\Rightarrow \sin C=\frac{10x/{{t}_{2}}}{x/{{t}_{1}}}\] \[\Rightarrow \sin C=\frac{10\,{{t}_{1}}}{{{t}_{2}}}\]\[\Rightarrow C={{\sin }^{-1}}\left( \frac{10\,{{t}_{1}}}{{{t}_{2}}} \right)\]
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