The angle of elevation of the top of a tower to a point on the ground is 61. At a point 600 feet farther from the base, in line with the bas

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Solution:

A. 574 (meh. full solution i guess...)

A right triangle is formed with the base measuring 600 ft. The adjacent angle is 32 degrees. Using the trigonometric function tangent, the height of the tower can be solved:
tan 32 = h/600

h = 573.98 ft

So I guess it's A...

The angle of elevation of the top of a tower to a point on the ground is 61 degrees. At a point 600 feet farther from the base, in line with the base and the first point and in the same plane, the angle of elevation is 32 degrees . Find the height of the tower and then choose the best answer.

A. 574

B. 503

C. 601

D. 414

The height of the tower is x.

\(tan(32°)=\frac{x}{\frac{x}{tan(61°)}+600ft}\)

   !

Using the rule  tan(angle)  =  opposite / adjacent ,  we can make two equations:

\(\tan(61^\circ)\ =\ \frac{h}{x}\\~\\ x\tan(61^\circ)\ =\ h\\~\\ x\ =\ \frac{h}{\tan(61^\circ)}\)

...and...

\(\tan(32^\circ)\ =\ \frac{h}{600+x}\\~\\ (600+x)\tan(32^\circ)\ =\ h\\~\\ 600+x\ =\ \frac{h}{\tan(32^\circ)}\\~\\ x\ =\ \frac{h}{\tan(32^\circ)}-600\)

Now we can equate both expressions of  x  and solve for  h:

\(\frac{h}{\tan(61^\circ)}\ =\ \frac{h}{\tan(32^\circ)}-600\\~\\ \frac{h}{\tan(61^\circ)}-\frac{h}{\tan(32^\circ)}\ =\ -600\\~\\ h(\frac{1}{\tan(61^\circ)}-\frac{1}{\tan(32^\circ)})\ =\ -600\\~\\ h\ =\ -600\div(\frac{1}{\tan(61^\circ)}-\frac{1}{\tan(32^\circ)})\\~\\ h\ \approx\ 573.5998\\~\\ h\ \approx\ 574\)

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