What is tan θ in this diagram?

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In summary, the conversation discusses a problem involving finding the value of tan θ in terms of two other angles with known tangents. Through various methods, including drawing a diagram and using trigonometric identities, it is shown that tan θ can be expressed as (y⋅sin α)/(x√2), where x and y are the sides of a right-angled triangle. The use of the cosine rule is found to be unnecessary in finding the solution.
  • #1
PsychoMessiah
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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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  • #2
Can you express tan theta in terms of two other angles you know the tangents for?
 
  • #3
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
PsychoMessiah said:
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
This does seem a bit long winded. Can you expand tan(a-b) directly in terms of tan a and tan b?
 
  • #6
sjb-2812 said:
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
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
PsychoMessiah said:
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.
 

1. What does tan θ represent in this diagram?

Tan θ, or tangent of θ, is a trigonometric function that represents the ratio of the length of the side opposite to the angle θ to the length of the adjacent side in a right triangle. It is calculated by dividing the length of the opposite side by the length of the adjacent side.

2. How do I find the value of tan θ in this diagram?

To find the value of tan θ, you need to know the measure of the angle θ and the lengths of the opposite and adjacent sides of the right triangle. Then, you can use the formula tan θ = opposite/adjacent to calculate the value.

3. Why is tan θ important in mathematics?

Tan θ is important in mathematics because it is a fundamental trigonometric function that is used to solve problems involving right triangles, angles, and sides. It also has many applications in fields such as physics, engineering, and astronomy.

4. Can tan θ be negative?

Yes, tan θ can be negative. The value of tan θ depends on the quadrant in which the angle θ lies. If θ is in the second or fourth quadrant, the value of tan θ will be negative. If θ is in the first or third quadrant, the value of tan θ will be positive.

5. What is the range of values for tan θ?

The range of values for tan θ is all real numbers, except for certain values where the tangent function is undefined (such as when the angle is a multiple of 90 degrees). This means that tan θ can take on any value between negative infinity and positive infinity.

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