Question

A suspension bridge has two main towers of equal height. A visitor on a tour ship

Answer 1

Answer:

The Height of the tower is 188.67 ft

Step-by-step explanation:

Given as :

The angle of elevation to tower = 15°

The distance travel closer to tower the elevation changes to 42° = 497 ft

Now, Let the of height of tower = h  ft

The distance between 42°  and  foot of tower = x  ft

So, The distance between 15° and  foot of tower =  ( x + 497 )  ft

So, From figure :

In Δ ABC

Tan 42° = $\frac{perpendicular}{base}$

Or , Tan 42° = $\frac{AB}{BC}$

Or,  0.900 = $\frac{h}{x}$

∴ h = 0.900 x

Again :

In Δ ABD

Tan 15° = $\frac{perpendicular}{base}$

Or , Tan 15° = $\frac{AB}{BD}$

Or,  0.267 = $\frac{h}{( x + 497 )}$

Or,  h = ( x + 497 ) × 0.267

So, from above two eq  :

     0.900 x =  ( x + 497 ) × 0.267  

Or, 0.900 x - 0.267 x =  497  × 0.267  

So, 0.633 x = 132.699

∴               x = $\frac{132.699}{0.633}$

Or,            x = 209.63  ft

So, The height of tower = h = 0.900 × 209.63

Or,                                      h = 188.67 ft

Hence The Height of the tower is 188.67 ft    Answer