Need geometry 40 points for the correct graphed as well! suppose points a(-13,7), b(5,6), and c(-8,-7) are the vertices of a triangle abc. graph the points and use the distance formula to find the perimeter of the triangle.

Refer to the table below if needed. second quadrant third quadrant fourth quadrant sin(1800- - cos(180° -) tan(180°-e) =- tane cot(1800-0) 10 it to solo 888 sin(180° +c) = - sine cos(180° +) =- cose tan(180° +c) = tane cot(180° +o) = cote sec(180° + c) = - seco csc(180° +2) = - csce sin(360° -) =- sine cos(360° -) = cose tan(360° - e) =- tane cot(360° -) = -cote sec(360° -) = seco csc(360° -) = csco sec(180° -) = csc(180° -) = csca 1991 given that sine = 3/5 and lies in quadrant ii, find the following value. tane

2y= 3 2 x−2 y = − 1 3 x + 1 y=− 3 1 x+1

If tan2a=cot(a+60), find the value of a, where 2a is an acute angle