A dive bomber has a velocity of 290 m/s at an angle θ below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.09 km. Find the angle θ.

Question

Y = Vy(T) + (1/2)gT^2 
X = Vx(T) 

where 

Y = distance of ground from point of release =2.15 km =2150 m 
Vy = vertical component of the initial velocity = 290*sin Θ 
Θ = angle of release with respect to the horizontal 
T = time for bomb to reach the target 
g = acceleration due to gravity = 9.8 m/sec^2 (constant) 
X = horizontal displacement of bomb from release point = 3.09 km = 3090 m 
Vx = horizontal component of initial velocity = 290(cos Θ) 

Substituting appropriate values, 

2150 = 290(sin Θ)(T) + (1/2)(9.8)T^2 

and simplifying, 

2150 = 290(sin Θ)(T) + 4.9(T^2) -- call this Equation 1 

For the horizontal component, 

3090 = 290(cos Θ)T 

Solving for "T"  

T = 3090/(cos Θ)(290) 

T = 10.6551/cos Θ 

and substituting this in Equation 1,  

2150 = 290(sin Θ)(10.65/cos Θ) + 4.9(10.65/cos Θ)^2 

I can't seem to simplify this? Can you help me?