Looking for θ in degrees.I was trying to see if I can use some trig identities to solve, but I'm still stuck.Thanks!. Created by pfmlml. science-mathematics-en - mathematics-en

-400cos(θ) + 200sin(θ) = -250divide by 50 -8cos(θ) + 4sin(θ) = -5 4sin(θ) = 8cos(θ) - 5squaring on both sides16sin^2(θ) = 64cos^2(θ) + 25 - 80cos(θ)=> 16(1 - cos^2(θ)) = 64cos^2(θ) + 25 - 80cos(θ)=> 16 - 16cos^2(θ)) = 64cos^2(θ) + 25 - 80cos(θ)=> 80cos^2(θ) - 80cos(θ) + 9 = 0solve for cosθ using quadratic formula and find cos inverse to get θ