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Look at the diagram geometry question

I know $\theta$, $D$ and ratio of $\frac{a}{b}$ (say $k$). I am trying to find out $\alpha$. Is this possible? I couldn't crack it. (I only know the ratio of $a$ to $b$, and I don't know $c$ as well)

UPDATE I think I have found a solution. Look at this.

x=tan($\theta$)*h

cos($\alpha$)=(2D-x)/a

cos($\alpha$)=2D/(a+b)

ratio of these two => (a+b)/a=2D/(2D-x)

We know a/b and D. We can find x. Since we know $\theta$, we can find $\alpha$.

I am starting a bounty for somebody who can prove this solution wrong. If it is correct, it is mine to keep.

Update2

Ok.. My bad.. I am overall decent guy but fall on my face on this one.

Community
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Frank
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    Would you be happy with a solution involving $c$? I think there is no solution not involving $c$. – Gerry Myerson Aug 25 '11 at 07:17
  • C is not known..The solution must be without c. – Frank Aug 25 '11 at 07:18
  • You can change $c$ and $\alpha$ together while not changing any of the other known values. There is no solution without $c$ unless there is more information that what is given. – anon Aug 25 '11 at 07:27
  • if we know D and theta then e can calculate all angles of big triangle with base on rectangular also all it''s sides – dato datuashvili Aug 25 '11 at 07:30
  • i think we dont need ratio a/b=k so start suppose theta is 30 degree it means that we know angle at the base of rectangular which is equal 90-30 =60(angle is arbitrary take any)then we know also angle at the top of rectangular side which is also 90-30=60 we can find adjacent angle by 180-(90-theta) and then make equation of two first sum of angle is 180 and second exterior angle is sum of it's non adjacent angles so you can find alpha – dato datuashvili Aug 25 '11 at 07:34
  • I am not so sure that proof is correct. – Frank Aug 25 '11 at 07:47
  • @Frank: I'm unable to parse that comment, but it can't work like that, a) because of what I wrote in my answer and b) because only $\theta$ is being used and you can obviously change $\alpha$ without changing $\theta$. – joriki Aug 25 '11 at 10:47
  • @joriki I agree, this is a problem with no solution. I will post another one, which is strangely is possible to solve and it is rather easy. stay tuned – Frank Aug 25 '11 at 11:26
  • @joriki, could there be a solution? I think I have found one. Let me hit it up. – Frank Aug 30 '11 at 11:24
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    Your alleged solution is unclear to me. Could you write out in equations what you sketched in words at the end? Especially the part "Since we know $\theta$, we can find $\alpha$". None of the equations you've written contains only $\alpha$ and no other unknowns. Note that the two new parameters $x$ and $h$ you introduced are left invariant by the procedure I described in my answer, so they can't determine $\alpha$, since the procedure changes $\alpha$. – joriki Aug 30 '11 at 12:01
  • Perhaps a good way for you to see that there can't be a solution is to imagine the line consisting of $a$ and $b$ as a rod that always moves along such that it passes through the corner on the left and the intersection of the vertical line labeled by $h$ with the second oblique line (the one at angle $\theta$). Then if you move the second oblique line and the bottom line upwards together (thus reducing $c$), all that changes is $\alpha$, $a$, $b$ and $c$, whereas $D$, $h$, $x$, $\theta$ and $a/b$ stay the same. – joriki Aug 30 '11 at 12:08
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    Frank: there probably is some (perhaps a real-world?) motivation behind this question and the other one you asked. It is unfortunate that you received a negative answer twice. I suggest that you ask a question describing the actual problem, maybe somebody can help you then and that would be much more satisfactory for everyone involved. Just my two cents. – t.b. Aug 30 '11 at 12:58
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    @theo you are right, there is a real world application. I will tell you guys when we ship. – Frank Aug 31 '11 at 00:26
  • @Frank: I'm not surprised :) Best wishes, – t.b. Aug 31 '11 at 00:32

3 Answers3

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Imagine changing $\alpha$ a bit. Now compensate for the change in $a/b$ by moving the top vertex up or down a bit. Now compensate for the change in $D$ by changing $c$ a bit. Now all you've changed is $\alpha$ and $c$, so you can't find $\alpha$ without $c$.

[Edit:]

Perhaps an easier way to see that there can't be a solution is to imagine the line consisting of $a$ and $b$ as a rod that always moves along such that it passes through the corner on the left and the intersection of the vertical line labeled by $h$ with the second oblique line (the one at angle $\theta$). Then if you move the second oblique line and the bottom line upwards together (thus reducing $c$), all that changes is $\alpha$, $a$, $b$ and $c$, whereas $D$, $h$, $x$, $\theta$ and $a/b$ stay the same.

joriki
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  • Changing $c$ will change $\alpha$ again, so this doesn't prove anything – Hannesh Aug 30 '11 at 12:23
  • @Hannesh: Perhaps that wasn't sufficiently clear. I meant changing $c$ by simply moving the bottom line up without changing any of the lines further up. I'll include my other explanation from the comments in the answer; perhaps that's more intuitive. – joriki Aug 30 '11 at 12:30
  • I will award the bounty to you Joriki as soon as I can.. thanks very much for your support – Frank Aug 31 '11 at 07:55
  • Deleted my comment as I noticed that Jiri had brought up the exact same point in his figure. – Jyrki Lahtonen Sep 04 '11 at 15:57
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You can compute $x$ and $h$. (BTW, your second equation should be $\cos(\alpha) = (D+x)/a$.). Even then, you cannot determine $\alpha$.

enter image description here

You just got the point $S$. According to the Intercept theorem all the red lines in the figure passing through $S$ have the same ratio $k = a/b$ but they have different angles $\alpha$ to the horizontal line.

EDIT. I am adding the explicit formulae for $x$ and $h$. We have $$ (1) \;\;\; \cos(\alpha) = (D+x) / a$$ $$ (2) \;\;\; \cos(\alpha) = 2D / (a+b)$$

From (1) and (2) we get

$$x = D (k-1) / (k+1) $$

Further, we obtain $h$ as $$ h = x / tan(\theta) $$

Now, the above geometric argument is applied. In addition, we see that $\alpha$ cannot be obtained from the equations (1) and (2) without knowing either $a$ or $a+b$.

Jiri Kriz
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2

I have copied your image below and added some constructions to help find $\tan(\alpha)$.

value-added copied image

Note that there are three similar right triangles, each with an angle of $\theta$. They are similar because they are right triangles and share the opposite (also called vertical) angles near the right arrow under $c\tan(\theta)$.

We can compute the length of the hypotenuse of the right triangle with the dashed, red side and angle of $\theta$ to be $D+c\tan(\theta)$, then compute the dashed, red side to be $D\cos(\theta)+c\sin(\theta)$. From there, it is pretty easy to compute $a$, $a+b$, and then take their ratio. Solving, we get $$ \tan(\alpha)=\cot(\theta)-\frac{a/b+1}{a/b}\frac{\cot(\theta)+c/D}{2} $$ This requires only that we know $a/b$, $c/D$, and $\cot(\theta)$, but $\tan(\alpha)$ is definitely dependent on $c/D$.

robjohn
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