it is required to measure the height of a tower.

I will rephrase the question for others to understand...hihihih

It is required to measure the height of a tower, $CB$, which is inaccessible. From point $A$, in the same horizontal plane with the base $C$, a right triangle $CAD$ is formed, and a horizontal line $AD$, $150$ feet in length, is measured. At $A$, the angle of elevation of the top of the tower is $32^o$, at D the angle of elevation is $28^o$. Find the height of the tower.

To solve, first draw the figure described above.....It looks like this:

Then, looking at $\triangle ABC$, we have...
$$BC = AC \tan 32^o$$

Then looking at $\triangle BCD$, we have...
$$BC = CD \tan 28^o$$

Then looking at $\triangle CAD$, we have...
$$CD^2 = AC^2+AD^2$$ $$\left(\frac{BC}{\tan 28^o}\right)^2 = \left(\frac{BC}{\tan 32 ^o}\right)^2 + 150^2$$ $$3.537BC^2 = 2.561BC^2 + 150^2$$ $$0.976BC^2 = 150^2$$ $$BC = 151.83$$

Therefore, we found that the height of tower $BC$ is $151.83$ feet....

Alternate solutions are encouraged......