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A flagpole is mounted on top of a tall building. At a distance of $250$m from the base of the building, the angles of elevation of the bottom and the top of the flagpole are $38^\circ$ and $40^\circ$ respectively. Calculate the height of the flagpole, correct to one decimal place. Ok so I know the answer is $14.5$m, but I don't know how to solve it or do the working out? Why does the angle of elevation have $2$ angles, can someone please explain this problem? Thank you.

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  • $\begingroup$ One angle of elevation is for the bottom of the flagpole, the second for the top of the flagpole. $\endgroup$ – peterwhy Aug 14 '13 at 12:53
  • $\begingroup$ I tried 250 Tan(40) but that didn't work, what's the formula for this type of question? $\endgroup$ – S.E. Chahine Aug 14 '13 at 12:57
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    $\begingroup$ $250 \tan 40^{\circ}$ is how high the top of the flagpole to the ground. $\endgroup$ – peterwhy Aug 14 '13 at 13:00
  • $\begingroup$ Try for understanding - there is no formula to life... $\endgroup$ – user1729 Aug 14 '13 at 13:08
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    $\begingroup$ life = woman + man + intercourse is a formula, which creates life $\endgroup$ – S.E. Chahine Aug 14 '13 at 13:09
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Define $\,b=$ height of building, $\,f=$height of flag, then basic trigonometry gives

$$\tan 38^\circ=\frac b{250}\implies b=250\tan38^\circ\\{}\\ \tan40^\circ=\frac{f+b}{250}\implies f+b=250\tan40^\circ$$

and thus we get that

$$f=250(\tan40^\circ-\tan38^\circ)=14.453$$

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    $\begingroup$ This is great, thank you! $\endgroup$ – S.E. Chahine Aug 14 '13 at 13:03
  • $\begingroup$ Is this the same to calculate the horizontal side and the hypotenuse? $\endgroup$ – S.E. Chahine Aug 14 '13 at 13:07
  • $\begingroup$ Nop. From the viewer's point, you are calculating the tangent of the elevation angle by means of the opposite leg (the building's or the builidng + flag's height) and the adyacent leg (the distance between the viewer and the building $\;=250\;$ meter) $\endgroup$ – DonAntonio Aug 14 '13 at 13:11
  • $\begingroup$ so what would be the formula for the other two sides? $\endgroup$ – S.E. Chahine Aug 14 '13 at 13:12
  • $\begingroup$ What "other two sides"? You have two straight-angle triangles: one with its hypotenuse to the base of the flag's pole and the other one with its hypotenuse to the pole's top. Both have the same horizontal leg $\,250\,$, and one has the other leg the building whereas the other one hast the building + the flag as its other leg. $\endgroup$ – DonAntonio Aug 14 '13 at 13:14
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Hint:

Since the top of flagpole has an angle of elevation of $40^{\circ}$ and the horizontal distance is 250 m, find out how far is the top of flagpole above ground. Similarly, find out how far is the bottom of flagpole above ground.

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Let the tall building with the flag pole be a segment lying on the y-axis in $\mathbb R^2$. Its base is the origin $(0,0)$. We denote by

$$P=(250,0)$$

the point whose distance from the bottom of the building is $250$ meters. We denote by $A$ the point corresponding to the bottom of the flagpole and by $B$ the one corresponding to the top of it, with

$$A=(0,y_1),$$ $$B=(0,y_2)$$

and $y_1<y_2$. The length of the flagpole is

$$l:=y_2-y_1.$$

Let us compute it; we know that

$$|AP|=\sqrt{250^2+y^2_1},$$ $$|BP|=\sqrt{250^2+y^2_2},$$

and

$$250=|AP|\cos(38^\circ),$$ $$250=|BP|\cos(40^\circ).$$

It follows that

$$250^2+y^2_1=\frac{250^2}{\cos^2(38^\circ)}\Rightarrow y_1=\sqrt{ \frac{250^2}{\cos^2(38^\circ) }-250^2}, $$ $$250^2+y^2_2=\frac{250^2}{\cos^2(40^\circ)}\Rightarrow y_2=\sqrt{\frac{250^2}{\cos^2(40^\circ)}-250^2}; $$

All you need now is to compute the difference

$$l=y_2-y_1=\sqrt{\frac{250^2}{\cos^2(40^\circ)}-250^2}- \sqrt{\frac{250^2}{\cos^2(38^\circ)}-250^2}. $$

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  • $\begingroup$ I think this process is unduly long and complex...after all it is basic trigonometry here. Yet it yields the correct answer. $\endgroup$ – DonAntonio Aug 14 '13 at 13:03
  • $\begingroup$ probably it would have been better with a nice picture of the building and so. $\endgroup$ – Avitus Aug 14 '13 at 13:07
  • $\begingroup$ I wonder why you chose not to use tangent. $\endgroup$ – peterwhy Aug 14 '13 at 13:09
  • $\begingroup$ nothing special behind my solution: I thought about it just like that... $\endgroup$ – Avitus Aug 14 '13 at 13:11
  • $\begingroup$ And $\frac{250^2}{\cos^2 40^{\circ}}-250^2 = 250^2 \sec ^2 40^{\circ} - 250^2 = 250^2 \tan ^2 40^{\circ}$ $\endgroup$ – peterwhy Aug 14 '13 at 13:13

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