Angles of Elevation: 173 metres from the base of the Eiffel Tower

michelleabellan

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Can anybody help with the below.

If I am stood at 173 metres from the base of the Eiffel Tower and the “angle of elevation” for me to look at the top of the tower is 60o what is the height of the tower?
 
Can anybody help with the below.

If I am stood at 173 metres from the base of the Eiffel Tower and the “angle of elevation” for me to look at the top of the tower is 60o what is the height of the tower?

First draw a sketch of the "situation" (assume your height is insignificant compared to that of the Eiffel tower)
 
Basically, you have a right triangle in which you are given one distance and one angle. You are given the angle 60 degrees and a "near side", from your eye to the base of the Eiffel Tower, of 173 m. You are asked to find the "opposite side", the height of the Eiffel Tower. "Near side" divided by "opposite side" is defined as the "tangent" of the angle. Calling the desired height "h", we have \(\displaystyle tan(60)= \frac{h}{173}\) so \(\displaystyle h= 173 tan(60)\).

Because this happens to be \(\displaystyle 60= \frac{180}{3}\) degrees this can be done without the direct use of trig functions! An equilateral triangle, by definition having all three sides of the same length, say "s", has all three angle the same, 180/3= 60 degrees. Drawing a line from one vertex perpendicular to the opposite side divides that side into equal parts, each of length s/2 and divides the vertex angle into two equal angle, each of 60/2= 30 degrees. Also, by the Pythagorean theorem, the altitude has length \(\displaystyle \sqrt{s^2- \frac{s^2}{4}}= s\sqrt{\frac{3}{4}}= \frac{s\sqrt{3}}{2}\).

Here \(\displaystyle \frac{s}{2}= 173\) so the height is \(\displaystyle 173\sqrt{3}\).

(Of course \(\displaystyle tan(60)= 1.7320= \sqrt{3}\).)
 
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