# Chapter 2 - Section 2.1 - Definition II: Right Triangle Trigonometry - 2.1 Problem Set: 15

Required trigonometric functions are- $\sin A = \frac{3}{5}$ $\\csc A = \frac{5}{3}$ $\\cos A = \frac{4}{5}$ $\\sec A = \frac{5}{4}$ $\\tan A = \frac{3}{4}$ $\\\cot A = \frac{4}{3}$

#### Work Step by Step

Steps to Answer- From the given diagram of triangle ABC, we will use given information and Pythagoras Theorem to solve for 'b'- We know that - $c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem) Therefore - $b^{2} =c^{2} - a^{2}$ $b^{2} = 10^{2} - 6^{2}$ $b^{2} = 100- 36$ $b^{2} = 64$ therefore $b = \sqrt 64$ = 8 Now we can write the required six T-functions of A using a = 6 , b = 8 and c = 10 $\sin A = \frac{a}{c} = \frac{6}{10} = \frac{3}{5}$ $\\csc A = \frac{c}{a} =\frac{10}{6} = \frac{5}{3}$ $\\cos A = \frac{b}{c} = \frac{8}{10} = \frac{4}{5}$ $\\sec A = \frac{c}{b} = \frac{10}{8} = \frac{5}{4}$ $\\tan A = \frac{a}{b} =\frac{6}{8} = \frac{3}{4}$ $\\\cot A = \frac{b}{a} = \frac{8}{6} = \frac{4}{3}$

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