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.

  • $\begingroup$ 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

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


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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


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.


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


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


Let us compute it; we know that

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


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

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

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

  • $\begingroup$ 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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