Dear Uncle Colin,

When I’m 8 metres away from a flagpole, the angle of elevation to its top is exactly 40º, according to this angle-measurey-majigger I have here. What will it be when I’m only five metres away?

The Hypotenuse Eludes One; Does One Look Into Trigonometric Expressions?

Hi, THEODOLITE, and thanks for your message!

This is a classical two-step trigonometry problem: the first step is to figure out whatever you can, and the second is to find the thing you need. (In principle - and certainly in the new GCSE - the first step might be several steps.)

Here, you can work out the height of the flagpole: using right-angled trigonometry on the bigger triangle (with a base angle of 40º and an adjacent side of 8m), it’s $8\tan(40º) \approx 6.713$ metres tall ((Adam Atkinson points out that, strictly, it’s 6.713 metres above the level at which it was measured, and that the whole question assumes horizontal ground.)).

Now move on to the smaller triangle, with unknown base angle, but an adjacent side of 5m and a height of 6.713 ((Or, if you use your calculator properly, Ans)) . You need to work out $\tan^{-1}\br{\frac{Ans}{5}} \approx 53.3º$.


“A ratio of 5:6.7 is about 15:20, so it’s close to a 3-4-5 triangle. The larger base angle there is 0.927 radians, or about 53º.”

“Thank you, sensei!”

Hope that helps!

* Edited 2018-08-22 to add a comment from Adam Atkinson. Thanks, Adam!