Answer
$\sin A = \frac{5}{6}$ , $\\csc A = \frac{6}{5}$
$\\cos A = \frac{\sqrt11}{6}$ , $\\sec A = \frac{6}{\sqrt 11}$
$\\tan A = \frac{5}{\sqrt 11}$ , $\\\cot A = \frac{\sqrt11}{5}$
Work Step by Step
Steps to Answer-
First we will draw the diagram of triangle ABC and then use the given information and Pythagoras Theorem to solve for 'b'-
We know that
$c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem)
therefore $b^{2} =c^{2} -a^{2}$
$b^{2} = 6^{2} - 5^{2}$
$b^{2} = 36 - 25$
$b^{2} = 11$
therefore $ b = \sqrt 11$
Now we can write the required six T-functions of A using a = 3, $b= \sqrt 7$ and c = 4
$\sin A = \frac{a}{c} = \frac{5}{6}$
$\\csc A = \frac{c}{a} = \frac{6}{5}$
$\\cos A = \frac{b}{c} = \frac{\sqrt 11}{6}$
$\\sec A = \frac{c}{b} = \frac{6}{\sqrt 11}$
$\\tan A = \frac{a}{b} = \frac{5}{\sqrt 11}$
$\\\cot A = \frac{b}{a} = \frac{\sqrt 11}{5}$
