Question

From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower. [Given ∶ √3 = 1.73]

Answer 1

The height of the building is AB = 7cm. 

∠EAD = 60° & ∠DAC = 45°. 

∵ ABCD is a rectangle. 

∴CD = AB = 7cm 

& AD = BC. 

Now,

In right angled triangle ADC, 

\(\frac{ED}{AD}\) = cot45° 

(∵∠DAC = 45°) 

⇒ AD = CD × 1 = 7cm. 

(∵ cot45° = 1 and CD = 7cm) 

Now, 

In right angled triangle EDA, 

\(\frac{ED}{AD}\) = tan60° 

(∵∠EAD = 60°) 

⇒ ED = AD × √3 

(∵ tan60° = √3) 

⇒ ED = 7√3m. 

(∵ AD = 7m) 

Now,

The height of the tower is CE = CD + DE 

= (7 + 7√3)m 

(∵ CD = 7m & ED = DE = 7√3m) 

=7(1 + √3)m 

=7(1 + 1.73)m 

(∵ √3 = 1.73) 

= 7 × 2.73m 

= 19.11m. 

∴ The height of the tower is 19.11m.