A) \[\frac{\sqrt{3}}{2}\]
B) \[\frac{5}{3}\]
C) \[\frac{2\sqrt{2}}{5}\]
D) \[\frac{4\sqrt{2}}{3}\]
Correct Answer: D
Solution :
[d] For T.I.R. \[45{}^\circ >C\] \[\therefore \sin 45{}^\circ >\operatorname{sinC}\] \[\therefore \frac{1}{\sqrt{2}}>\frac{4/3}{n}\] \[\therefore n>\frac{4\sqrt{2}}{3}\]You need to login to perform this action.
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