My guess is that you're doing the Law of Cosines? You have everything you need for that except the angle theta, which is the thing you need to find. It's set up like this: (8)^2 = (10)^2 + (5)^2 -[2(10)(5)cos A] I used A instead of theta. Doing that math, you have: 64 = 100 + 25 -[ 100 cos A]; 64 = 125 - 100 cos A;
-61 = - 100 cos A; -61 / -100 = cos A; .61 = cos A. Now use your inverse function on your calculator to find cos^-1(.61) and that equals 52.4
2b^2. This is because you have a positive a^2 and a negative so they cross it one another so it just leaves the b’s.
Hello
the line passes by A ( 0 ; -2) and B (5 ; 0) equation : y = ax+b
a is a slop : a = (yb - ya)/(xb -xa)
a = (0+2)/(5-0)
a =2/5
y = (2/5)x+b
if x=5 and y= 0
0 = (2/5)(5)+b
b= - 2
the equation is : y = (2/5)x-2