+0  
 
0
761
1
avatar

In making a topographical map, it is not practical to measure directly heights of structures such as mountains. This exercise illustrates how some such measurements are taken. A surveyor whose eye is a = 7 feet above the ground views a mountain peak that is c = 4 horizontal miles distant. (See the figure below.) Directly in his line of sight is the top of a surveying pole that is 10 horizontal feet distant and b = 8 feet high. How tall is the mountain peak? Note: One mile is 5280

eet. 

 Mar 13, 2018
 #1
avatar+37146 
0

Using similar triangles:

 

1 ft / 10 ft  =   x ft / (5280 x 4)ft      x = 2112   PLUS the 7 ft observer's 'eye-height' = 2119 ft high mountain

 

 

Using trig fxns:

For the surveyor's pole:

    Tan theta = 1/10     theta = 5.71 degrees

 

For the mtn:

    tan 5.71  = opp/adj  = opp/(4 x 5280 ft)    opp = 2112     2112 + observer's height (7 ft .....very tall surveyor) = 2119 ft high

 Mar 13, 2018

2 Online Users

avatar