Story problem from pre-calc.Two buildings of equal height are 800 feet apart. An observer on the street between the buildings measures the angles...

Let the nearest distance of the observer from one of the two equally high buildings be x.

The angle of the observing equal height from the nearer point should be higher than that of distant building. So the nearest angle of observation is 41 deg.

Now Let A be the position of the observer, B the feet of the nearest building and C the top most point of the nearest building.

Now ABC is a right angled triangle, with angle ABC=90 deg and angle BAC =41 deg.

Therefore, tan BAC= BC/AB = h/x, where h is the hight.

Therefore, x= h/tan(angleBAC) = h/tan41................(1)

Similarly, the distance to the feet of the other building is 800-x = h/tan27deg............................................(2)

(1)/(2) eliminates h :

x/(800-x) =tan27/tan41 or

x = (800-x)tan27/tan41 or

x(1+tan27/tan41)=800tan27/tan41.

x=800 tan27/(tan41+tan27)=295.6315 feet is the nearest distance of the observer.

The distance between the feet of the other building and the observer is 800-295.6315 feet = 504.3685 feet.

Now, the other unnown is h, the hight of the building.

By equation (1) h=xtan41 = 295.6315*tn41 = 256.9886 feet.