The tower is 700 meters high. Suppose a building is erected such that the base of the building is on the same plane as the base of the tower, the angle of elevation from the top of the building to the top of the tower is 80.38° and the angle of depression from the top of the building to the foot of the tower is 65.05°. How high would the building have to be?

Have a look at the (very quick) picture I drew of the situation. We have the tower (T) and the building (B). The tower is 700m tall and the angles of elevation and depression as described in the question are shown in the picture.

You'll notice that (blue) triangles can be formed when those angles are sketched in. There's a big scalene triangle, △ABC, and two smaller right-angled triangles, △ADB and △BDC. The question wants the height of the building, which corresponds to the green line (DC). We have sides and we have angles, so sounds like a job for trigonometry.

We know that AC = 700. Therefore, AD + DC = 700, so AD = 700 - DC (1). We have 1 equation but 2 unknowns, so we need to develop a second equation so we have a solvable system. We can get that second equation using the small triangles:

For △ADB: tan80.38 = AD/BD → BD = AD/tan80.38 (2)

For △BDC: tan65.05 = DC/BD → BD = DC/tan65.05 (3)

Yes a third variable (BD) has been introduced, but notice that it appears in both equations and we can rearrange them to isolate it. This is why I used tan instead of cos or sin - tan is the only one of the 3 that shares that side for both triangles. After the rearranging we get equations (2) and (3), which can be set equal to each other to get rid of that third variable:

AD/tan80.38 = DC/tan65.05 (4)

We can now sub (1) into (4) and solve for DC:

(700 - DC)/tan80.38 = DC/tan65.05

(700 - DC)tan65.05 = DCtan80.38

700tan65.05 - DCtan65.05 = DCtan80.38

DCtan80.38 + DCtan65.05 = 700tan65.05

DC(tan80.38 + tan65.05) = 700tan65.05

DC = 700tan65.05/(tan80.38 + tan65.05)

DC = 187.25