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# I NEED HELP (COLLEGE ALGEBRA)

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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 = 6 feet above the ground views a mountain peak that is c = 5 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 = 7 feet high. How tall is the mountain peak? Note: One mile is 5280 feet.

_________ft

idenny  Feb 21, 2017
edited by idenny  Feb 21, 2017
edited by idenny  Feb 21, 2017
edited by idenny  Feb 21, 2017

#2
+18381
+20

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 = 6 feet above the ground views a mountain peak that is c = 5 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 = 7 feet high.

How tall is the mountain peak?

Note: One mile is 5280 feet.

_________ft

$$\begin{array}{|rcll|} \hline \dfrac{H-a}{c} &=& \dfrac{b-a}{10} \\ H-a &=& c\cdot \left( \dfrac{b-a}{10} \right) \\ H &=&a+ c\cdot \left( \dfrac{b-a}{10} \right) \quad & | \quad a = 6\ feet, \quad b = 7\ feet, \quad c = 5\ miles\cdot \frac{5280\ feet}{mile} \\\\ H &=&6+ 5\cdot 5280 \left( \dfrac{7-6}{10} \right) \\\\ H &=&6+ 5\cdot 528 \\\\ H &=&2646\ feet \\ \hline \end{array}$$

the mountain peak is 2646 feet tall.

heureka  Feb 22, 2017
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#1
+4174
+2

First we need to find the length of the extra piece of triangle that isn't drawn on the picture, the part that is behind the man. Let's call that part "x." Two similar triangles are formed. The first one is formed by the man, the line of sight, and the ground, x. The second one is formed by the surveying pole, the line of sight, and the ground, x+10. Since these triangles are similar, the height of the man over the length of his ground equals the height of the pole over the length of its ground.

$$\frac{6}{x}=\frac{7}{x+10}$$

6(x+10)=7x

6x+60=7x

60=x

Now that we know x, we can make two different similar triangles. The first one is formed by the mountain, the line of sight, and the ground, c+60. The second is formed by the man, the line of sight, and the ground, 60.

So

$$\frac{m}{c+60}=\frac{6}{60}$$

Where "m" is the height of the mountain. Before we put "c" in, we have to convert it to feet. 5 miles = 5(5280) ft = 26400 ft.

$$\frac{m}{26400+60}=\frac{6}{60}$$

$$\frac{m}{26460}=\frac{6}{60}$$

m = 0.1(26460) = 2,646 ft

hectictar  Feb 21, 2017
#2
+18381
+20

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 = 6 feet above the ground views a mountain peak that is c = 5 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 = 7 feet high.

How tall is the mountain peak?

Note: One mile is 5280 feet.

_________ft

$$\begin{array}{|rcll|} \hline \dfrac{H-a}{c} &=& \dfrac{b-a}{10} \\ H-a &=& c\cdot \left( \dfrac{b-a}{10} \right) \\ H &=&a+ c\cdot \left( \dfrac{b-a}{10} \right) \quad & | \quad a = 6\ feet, \quad b = 7\ feet, \quad c = 5\ miles\cdot \frac{5280\ feet}{mile} \\\\ H &=&6+ 5\cdot 5280 \left( \dfrac{7-6}{10} \right) \\\\ H &=&6+ 5\cdot 528 \\\\ H &=&2646\ feet \\ \hline \end{array}$$

the mountain peak is 2646 feet tall.

heureka  Feb 22, 2017

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