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.

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?

To simplify the equation, we can start by multiplying throughout by (cos Θ):

2150cos Θ = 290sin Θ * 10.6551 + 4.9 * (10.6551)^2

Next, we can use the trigonometric identity sin 2Θ = 2sin Θ * cos Θ to rearrange the equation:

2150cos Θ = 20.6551 * 290sin Θ * cos Θ + (4.9 * 10.6551^2) * (2sin^2 Θ - 1)

Expanding further:

2150cos Θ = 20.6551 * 290sin Θ * cos Θ + (4.9 * 10.6551^2) * 2sin^2 Θ - (4.9 * 10.6551^2)

Rearranging the terms:

2150cos Θ - 20.6551 * 290sin Θ * cos Θ - (4.9 * 10.6551^2) * 2sin^2 Θ + (4.9 * 10.6551^2) = 0

Now, we can factor out common terms:

(2150 - 20.6551 * 290cos Θ) * cos Θ - (4.9 * 10.6551^2) * 2sin^2 Θ + (4.9 * 10.6551^2) = 0

Simplifying further:

2150cos Θ - 5984.921cos Θ + 104853.624sin^2 Θ - 104853.624 = 0

Combining like terms:

-3834.921cos Θ + 104853.624sin^2 Θ - 104853.624 = 0

We can further simplify by dividing throughout by -104853.624:

0.03659cos Θ - sin^2 Θ + 1 ≈ 0

Now, we have a quadratic equation in terms of sin Θ and cos Θ. However, we can make this equation easier to work with by using the identity sin^2 Θ = 1 - cos^2 Θ:

0.03659cos Θ - (1 - cos^2 Θ) + 1 ≈ 0

0.03659cos Θ - 1 + cos^2 Θ + 1 ≈ 0

Simplifying further:

0.03659cos Θ + cos^2 Θ ≈ 0

Now, we can solve this equation using numerical methods or by graphing the equation and finding the values of cos Θ that give a solution.