What is tan θ in this diagram?

  • #1
a41lrt.png


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It's supposed to be a simple problem. But I can't for the life of me figure out how to go about it. I managed to find out cos θ using the cosine rule, but it is a very long expression and looks to be going in a direction opposite of the solution. cos θ is (2x^2 + 2xy + y^2 + x*sqrt(2) - y) / (2 * (2x^2 + 2xy + y^2) * (x*sqrt2)).

Any help on this would be appreciated.
 
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Answers and Replies

  • #2
445
5
Can you express tan theta in terms of two other angles you know the tangents for?
 
  • #3
Filip Larsen
Gold Member
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Perhaps you can form a right-angled triangle involving theta and the two sides? Hint: try draw the full rectangle and see what that brings you. If you can't use angle theta directly then perhaps some other angle easily derived from it ...
 
  • #4
1,184
223
a41lrt.png


It's supposed to be a simple problem. But I can't for the life of me figure out how to go about it. I managed to find out cos θ using the cosine rule, but it is a very long expression and looks to be going in a direction opposite of the solution. cos θ is (2x^2 + 2xy + y^2 + x*sqrt(2) - y) / (2 * (2x^2 + 2xy + y^2) * (x*sqrt2)).

Any help on this would be appreciated.
From intersection point of pieces "x" and "y" put a normal "n" to a hypotenuze of a big triangle. Then you have:
n : x√2 = sin θ , n : y = sin α
From this you have: sin θ = (y⋅sin α)/(x√2)
Knowing that sin2α = x2/(x2+(x+y)2) and that 1+ctg2θ = 1/sin2θ , you should obtain correct result (A).
 
  • #5
445
5
This does seem a bit long winded. Can you expand tan(a-b) directly in terms of tan a and tan b?
 
  • #6
1,184
223
This does seem a bit long winded.
Untitledd9181.png

qrt{2}}\cdot%20\sin%20\alpha%20%3D\frac{y}{\sqrt{2}\cdot%20\sqrt{x^{2}&plus;%28x&plus;y%29^{2}}}.gif

^{2}%29}{y^{2}}-1%3D\frac{4x^{2}&plus;4xy&plus;y^{2}}{y^{2}}%3D\frac{%282x&plus;y%29^{2}}{y^{2}}.gif


1 min for drawing, 2 min for calculation, 3 min for Latex. This is how long it takes when derived from first principles.
 
  • #7
Filip Larsen
Gold Member
1,300
222
By inspection of the diagram one can establish ##\tan(\pi/4-\theta) = x/(x+y)## from which it is easy to expand and solve for ##\tan(\theta)## (but here left as an exercise for the original poster).
 
  • #8
Thanks for answering, everyone.

zoki85, that is very neatly done. Turns out we didn't need the cosine rule at all.

Filip Larsen, yes it is established that tan (45 - θ) = x / ( x + y ). But after expansion, we are left with (1 - tan θ) / (1 + tan θ ) using this formula...

idents07.gif


If you have an answer in mind, please share it.

Thanks.
 
  • #9
34,331
5,975
Thanks for answering, everyone.

zoki85, that is very neatly done. Turns out we didn't need the cosine rule at all.

Filip Larsen, yes it is established that tan (45 - θ) = x / ( x + y ). But after expansion, we are left with (1 - tan θ) / (1 + tan θ ) using this formula
We are not "left with" (1 - tan θ) / (1 + tan θ ) -- we are left with an equation whose right side is this. Write the whole equation and solve it for tan θ.
PsychoMessiah said:
idents07.gif


If you have an answer in mind, please share it.

Thanks.
 
  • #10
Right. Get it now. Thanks.
 

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