Question

The angle of elevation of the top of a cable tower when
observed from the top of a 9 m tall building is 60° and the angle of depression of the foot of this tower is 45°. Determine the height of the cable tower.​​

Answer 1

Step 1: Draw the figure
Represent all the angles and height in the diagram

Step 2: Find AD

𝐀𝐁 = 𝐃𝐂 = 𝟗 𝐦
In ∆𝑨𝑫𝑪
𝐭𝐚𝐧 𝟒𝟓° = $\frac{DC}{AD}$
𝐭𝐚𝐧 𝟒𝟓° = $\frac{9}{AD}$
𝟏 = $\frac{9}{AD}$
AD = 9 m

Step 3: Find height of the cable tower

Height of tower = EC
𝑬𝑪 = 𝑬𝑫 + 𝑫𝑪
In ∆𝑨𝑫𝑬
𝐭𝐚𝐧 𝟔𝟎° = $\frac{ED}{AD}$
$\sqrt{3}$ = $\frac{ED}{9}$
𝑬𝑫 = 9$\sqrt{3}$ 𝒎
𝑬𝑪 = 𝑬𝑫 + 𝑫𝑪
𝑬𝑪 = 9$\sqrt{3}$ + 𝟗 = 9($\sqrt{3}$+1) = 24.57 𝒎

Height of the cable tower is 24.57 m