If A and B are 100 ft apart find the height of tower

Question

A tower stands on level ground and is due north point A. The tower is also due west of another point B. From the top of the tower and angles of depression of A and B are 60° and 45° respectively. If A and B are 100 ft apart find the height of tower

Answer 1

Let h be the tower's height and point C and D be the base and top of the tower respectively.

 

∠CAD is 60° so |AC| = h tan 60° = h√3

 

∠CBD is 45° so |BC| = h tan 45° = h

 

then use Pythagoras's theorem: should yield h = 50

Answer 2

Let h be the tower's height and point C and D be the base and top of the tower respectively.

 

∠CAD is 60° so |AC| = h cot 60° = h / √3

 

∠CBD is 45° so |BC| = h cot 45° = h

 

AC ⊥ BC and |AB| = 100 so use the Pythagorean Theorem.

 

|AC|² + |BC|² = |AB|²

 

(h / √3)² + h² = 100²

 

h² / 3 + h² = 10000

 

4h² / 3 = 10000

 

h² = 30000/4 = 7500 = 3·50²

 

h = 50√3 ≈ 86.6 feet