A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes the surface of the water 2.34 102 m from the shore. A building stands on the shore, and the laser beam hits a target at the top of the building. The goal is to find the height of the target above sea level.

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

[tex]tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2[/tex]

[tex]\hat{ECD} = tan^{-1} 2 = 63.43^o[/tex]

This is also the angle ACB, so we can find the length AB:

[tex]tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}[/tex]

[tex]2 = \frac{AB}{234}[/tex]

[tex]AB = 2*234 = 468 m[/tex]

So the height of the building is 468m