We have a rectangle ABCD, and a point P on the diagonal AC. From P we see BC at an angle $\alpha$. Knowing $\alpha$, AP and AB, find BC.

Here's a probably unnecessary illustration of the problem, with known data in red:

Rectangle ABCD

I am lost with this, certainly, extremely simple problem. I might use the cosine law to define BC as a function of $\alpha$, PC and PB, but I find no way of finding those latter. I thought of finding PB constructing a parallelogram including sides PB and AP, having BC as diagonal, but I'm stuck there since I, in my poor knowledge on solving parallelograms, I would need the other diagonal of the parallelogram to find PB. On the other hand, I looked where could I transpose $\alpha$ as to express its trigonometric functions using known information, but didn't found anything useful.

Sorry for coming to you with high-school problems (I feel ridiculous to be asking this after three semesters of real analysis, one of complex, and one of linear algebra). I'd appreciate, in addition to a method of resolution, ressources that might help me get a comprehensive approach to planar, high-school like geometry.


1 Answer 1

  1. Find $\angle APB = 180º - \alpha$.
  2. Use the sine rule in $\Delta APB$ to find $\angle PBA$.
  3. Find $\angle PAB = 180º - \angle APB - \angle PBA$.

then since $\Delta CAB$ is right-angled, $\tan \angle PAB = \frac{BC}{AB}$, so you now have $BC$.

  • 2
    $\begingroup$ A shortcut: once you know $\angle PAB,$ you can write $BC = AB \tan\angle PAB$. $\endgroup$
    – David K
    Feb 8, 2021 at 0:21
  • $\begingroup$ Thanks for the shortcut! It was really nice of you to point that out. $\endgroup$
    – Toby Mak
    Feb 8, 2021 at 0:25
  • $\begingroup$ Dobby is a free elf! Really, I was haunted by the idea of not being able to solve this... I clearly need to practice more problems needing the use of the laws of sines and cosines. $\endgroup$
    – Albert
    Feb 8, 2021 at 0:51
  • $\begingroup$ This solution isn't elegant, but I looked and couldn't find one (an elegant solution), although I'd be surprised if one didn't exist. $\endgroup$ Feb 8, 2021 at 1:15

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