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I have a good understanding of the basic SohCahToa trig functions, but this kinda stumped me, since I don't have two parts of the information that is needed, here is a example image: https://i.imgur.com/ljztgBm.png

I have though of several methods, such as using an angle function between H and X, and a few others, but all seem too hacky, if anyone has a better idea, I'd really appreciate it :-) least number of steps the better

If I had a guess, the answer would involve X and H being used to find the length of Y

known variables: A,X,H,P1-4, A is always vertical, the pink is always a rectangle

desired information: the length of Y, which is the width of the pink rectangle

Andrew
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Weeve Ferrelaine
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2 Answers2

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In the triangle $P_1P_2P_4$, we have $$\cos(\widehat{P_2P_1P_4})=\frac{x^2+H^2-A^2}{2xH}$$

In the pink right triangle, $$\cos(\widehat{P_2P_1P_4})=\frac{Y}{H}$$

Can you deduce $Y$?

user5402
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Let the left top corner of the rectangle be P5. Call a=|P5-P3| (distance between P5 and P3) and b = |P5-P1|. Then

$$H^2 = b^2 + (X+a)^2$$

$$A^2 = b^2 + a^2 $$

Solve the second one for b, i.e $b^2 = A^2 - a^2$. Then put this into the first equation:

$$H^2 = A^2 -a^2 + X^2 + 2aX + a^2$$

Then

$$a = \frac{H^2 - A^2 - X^2}{2X}$$

Then $Y = X + a = X + \frac{H^2 - A^2 - X^2}{2X} = \frac{X^2 + H^2 - A^2 }{2X}$

Sina
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  • Could you give some articles I could read that proofed this example? I would like to learn from answers as well, and not just to use them :-) – Weeve Ferrelaine Apr 30 '14 at 17:58
  • What do you mean article? I wrote this proof now and it only uses the fact that if you have right triangle with sides a,b,c where c is the hyp. then c^2 = a^2 + b^2. Ask me any part that is not clear. In the answer the triangle P1P4P5 is a right triange therefore the first equation and P3P1P5 is a right triangle therefore the second equation. – Sina Apr 30 '14 at 18:01
  • Alright, thats what I wanted, thankyou, will be checking your work now :D – Weeve Ferrelaine Apr 30 '14 at 18:04
  • I understand your post by about nine tenths now, and hit a small roadblock, and realized that X+Dot((P3-P1),(P2-P1)) would yeild the same results (I have a much better background in linear, sowwy), although your answer looks legit enough to have worked, I will mark it incase anyone else needs to read this post – Weeve Ferrelaine Apr 30 '14 at 18:38