Answer
Coordinates of point B-
(8,6)
Requires T-functions of A-
$\sin A = \frac{3}{5}$
$\\cos A = \frac{4}{5}$
$\\tan A = \frac{3}{4}$
Work Step by Step
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
$\sin A = \frac{a}{c} = \frac{6}{10} = \frac{3}{5}$
$\\cos A = \frac{b}{c} = \frac{8}{10} = \frac{4}{5}$
$\\tan A = \frac{a}{b} = \frac{6}{8} = \frac{3}{4}$
