Answer:
[tex]Tan A = \frac{7}{5} \\[/tex]
Consider right angle triangle ABC,∠B= 90°.
[tex]tan = \frac{Opposite}{Adjacent} \\\\We \ get \ opposite = 7 \ and \ adjacent = 5\\hypotenuse^{2}= opposite^{2}+adjacent^{2}\\\\[/tex]
[tex]= 49+25 = 74\\hypotenuse^{2} = 74\\hypotenuse = \sqrt{74} \\\\Sin A = \frac{Opposite}{hypotenuse} = \frac{7}{\sqrt{74} }[/tex]
[tex]cosec A = \frac{1}{sinA} = \frac{1}{\frac{7}{\sqrt{74} } } = \frac{\sqrt{74} }{7}[/tex]
