Answer to Question #133454 in Trigonometry for Joseph Se

Question

An observer wishes to determine the height of a tower. He takes sights at the top of the tower from A and B, which are 50 feet apart, at the same elevation on a direct line with the tower. The vertical angle at point A is 300 and at point B is 400. What is the height of the tower?

Answer 1

Let the Height of tower be h,

Distance between point B and tower be x

When the observer is looking from point A

tan30$\degree$ =$\frac{h}{50+x}$

$50+x=1.732h$

x=1.732h-50...(1)

Now observer is looking from Point B

tan40$\degree$ =$\frac{h}{x}$

h=x$\times$ tan40$\degree$

=(1.732h-50)$\times 0.839$

h=1.4533h-41.9549

h-1.4533h=-41.9549

-.04533h=-41.9549

h=$\frac{41.9549}{0.4533}$

h=92.55feet

Hence the height of tower is 92.55 feet