## Trigonometry 7th Edition

Coordinates of point B- (8,6) Requires T-functions of A- $\sin A = \frac{3}{5}$ $\\cos A = \frac{4}{5}$ $\\tan A = \frac{3}{4}$
Steps to solution- Given that angle A is in standard position- Length AC = 8 Therefore x-coordinate of point B is '8' as it is the distance traveled along positive direction of x-axis. Length BC = 6 Therefore y-coordinate of point B is '6' as it is the distance traveled along positive direction of y-axis. Hence (8,6) are the coordinates of point B. Solution for T-functions- Considering triangle ABC right angled at C BC = 6 = a AC = 8 = b (concluded from above solution) Using this data and Pythagoras Theorem to solve for 'c'- $c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem) $c^{2} = 6^{2} + 8^{2}$ $c^{2} = 36 + 64$ $c^{2} = 100$ Therefore c = 10 Now we can write the asked T-functions of A using a = 6, b=8 and c = 10 $\sin A = \frac{a}{c} = \frac{6}{10} = \frac{3}{5}$ $\\cos A = \frac{b}{c} = \frac{8}{10} = \frac{4}{5}$ $\\tan A = \frac{a}{b} = \frac{6}{8} = \frac{3}{4}$