A) \[\frac{4}{\sqrt{65}}\]
B) \[\frac{4}{7}\]
C) \[\frac{7}{4}\]
D) \[\frac{\sqrt{65}}{7}\]
Correct Answer: A
Solution :
| In right \[\Delta TSP,\] \[\sin y{}^\circ =\frac{Perpendicular}{Hypotenuse}=\frac{TS}{PT}=\frac{4}{\sqrt{65}}\] |
| So, option [a] is correct. |
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