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

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Guest Mar 13, 2018

#1**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

ElectricPavlov Mar 13, 2018