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 turned, 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 degrees, at D the angle of elevation is 28 degrees. find the height of the tower

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

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......