Answer:
θ = 62.72°
Explanation:
The projectile describes a parabolic path:
The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).
The equation of uniform rectilinear motion (horizontal ) for the x axis is :
x = x₀+ vx*t   Formula (1)
vx = v₀x 
Where:  
x: horizontal position in meters (m)
x₀: initial horizontal position in meters (m)
t : time (s)
vx: horizontal velocity  in m/s 
v₀x: Initial speed in x  in m/s 
The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:
y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)
vfy= v₀y -gt Formula (3)
Where:  
y: vertical position in meters (m)  
y₀ : initial vertical position in meters (m)  
t : time in seconds (s)
v₀y: initial  vertical velocity  in m/s  
vfy: final  vertical velocity  in m/s  
g: acceleration due to gravity in m/s²
Data
v₀ = 120 m/s  , at an angle  θ above the horizontal
v₀x= 55 m/s
x-y components of the initial  velocity ( v₀)
v₀x = v₀*cosθ Equation (1)
v₀y = v₀*sinθ   Equation (2)
Calculating of the angle θ 
We replace data in the  Equation (1)
55 =  120*cosθ
cosθ = 55 / 120

θ = 62.72°