Let Theta=A. The given equation can be written: -18cosA-15tanA=-20secA ··· Eq.1
tanA=sinA/cosA and secA=1/cosA So that, Eq.1 can be rewritten as follows:
-18cosA-15sinA/cosA=-20/cosA Multiply both sides by cosA.
-18cos²A-15sinA=-20 ··· Eq.2 Use the identity: cos²A=1-sin²A, and plug this into Eq.2.
-18(1-sin²A)-15sinA=-20 We have:18sin²A-15sinA+2=0 ··· Eq.3
Let sinA=t Eq.3 can be rerwitten as follows:
18t²-15t+2=0 Factor the equation.
(6t-1)(3t-2)=0 We have: t=1/6, or 2/3
That is: sinA=1/6, or 2/3 (0<a<360°) So, A=arc sin(1/6), or arc sin(2/3)
And sinA=sin(180-A) So, (180-A)is the answer as well.
Therefore the answer is: Theta=approx 9.594°, 41.810°, 170.406°, or 138.190°