Answer:
Step-by-step explanation:
[tex]sec A=\frac{14}{\sqrt{171} } \\cos ~A=\frac{\sqrt{171} }{14} \\sin ~A=\sqrt{1-cos^2A} =\sqrt{1-\frac{171}{196} } =\sqrt{\frac{196-171}{196} } =\sqrt{\frac{25}{96} } =\pm\frac{5}{14} \\csc A=\pm \frac{14}{5}[/tex]
as A is in quadrant 1
[tex]so~csc~A=\frac{14}{5}[/tex]
in quadrant 1 both sin A and csc A are positive.
