Set up two separate equations in two different variables, t and Θ, the first based on what's happening horizontally and the 2nd based on what's happening vertically. (This is super common approach in kinematics.)
First, break down the bullet's initial vector into its 2 components: 70 cosΘ in the x direction, 70 sinΘ in the y.
Next, set up an equation for the x direction: (70cosΘ)⋅t = 180 (we want its initial velocity times time to = dist.)
Then set up an eqn in the y direction: -4.9t2 + (70sinΘ)⋅t = 0 (we want the bullet to be at height 0 after t secs)
Isolate t in the 1st to get t = 18/(7cosΘ) then substitute that expression for t in the 2nd equation, after first factoring out a t.
We get 18/(7cosΘ)(-4.9(18/(7cosΘ)) + 70sinΘ) = 0
We need a calculator to solve -4.9(18/(7cosΘ)) + 70sinΘ = 0 but doing so gives Θ ~ .18413395 rads
We can verify this answer by substituting that angle into the x eqn to get t ~ 2.61564 secs and then plugging that t and that Θ into the second to confirm it will ~ 0.
