First thing: Draw a diagram. You need it to help visualise the problem. Continue reading once you have a diagram:

This problem involves RIGHT-ANGLED triangles. Whenever you are asked to deal with right-angled triangles, always think of the word PYTHAGORAS, and then think of SIN, COS & TAN.

Since this problem involves angles, we will use sin, cos & tan.

Consider the smaller right-angled triangle with the 37 degree angle:

The side OPPOSITE this angle is H (the height of the building / the vertical distance from the ground to the bottom of the antenna). Let's call the distance from the point on the ground to the bottom of the building X. This is the side ADJACENT to the 37 degree angle.

We have the OPPOSITE & the ADJACENT so what will we use? Tan's the man for the job.

Tan(37) = H/G

Now let's consider the larger right-angled triangle with the 51 degree angle:

Again, the adjacent side is G. The opposite side is the vertical distance from the ground to the top of the antenna, which is the height of the building plus the height of the antenna i.e. H + 100

Tan(51) = (H+100)/G

So now we have 2 equations with 2 variables (H & G): Tan(37) = H/G & Tan(51) = (H+100)/G

What are we looking for? H. So let's get rid of G: G = H/Tan(37)

Now let's plug this into the second equation: Tan(51) = (H+100)/(H/Tan(37))

Invert & multiply to tidy up the ugly fraction: Tan(51) = (HTan(37)+100Tan(37))/H

Now cross-multiply: HTan(37) + 100Tan(37) = HTan(51)

Bring the Hs to one side: H(Tan(51) - Tan(37)) = 100Tan(37)

Therefore, H = 100Tan(37)/(Tan(51) - Tan(37)) = 156.55

156.55 inches? centimetres? ALWAYS REMEMBER YOUR UNITS: H = 156.55 ft