Find the length of a triangle's side
$begingroup$
I have the following triangle and I'm supposed to find the value of x.
First thought that came to mind is to use the the following tangent equation
$$tan(y)=frac{x}{27}$$ and $$tan(19+y) = frac{2x}{27}$$ which implies that $$tan(19+y) =2tan(y)$$ and solve for $y$ and once I've found $y$ I can easily find $x$.
I don't have an easy solution for the last equation and I feel that I'm complicating things unnecessarily. (It's a grade 10 geometry question).
Edit: Using the fact $tan(a+b) = dfrac{tan(a)+tan(b)}{1-tan(a)tan(b)}$ yields the following quadratic equation $$2tan(19)tan^2(y)-tan(y)+tan(19) = 0$$which gives you that $tan(y) = 0.890833$ or $tan(y)= 0.561272$ which means $x = 27times0.890833$ or $x= 27times 0.561272$.
So the question is now which $x$ should I pick and why? and is this more complicated than it should be?
geometry proof-verification proof-explanation triangle alternative-proof
$endgroup$
add a comment |
$begingroup$
I have the following triangle and I'm supposed to find the value of x.
First thought that came to mind is to use the the following tangent equation
$$tan(y)=frac{x}{27}$$ and $$tan(19+y) = frac{2x}{27}$$ which implies that $$tan(19+y) =2tan(y)$$ and solve for $y$ and once I've found $y$ I can easily find $x$.
I don't have an easy solution for the last equation and I feel that I'm complicating things unnecessarily. (It's a grade 10 geometry question).
Edit: Using the fact $tan(a+b) = dfrac{tan(a)+tan(b)}{1-tan(a)tan(b)}$ yields the following quadratic equation $$2tan(19)tan^2(y)-tan(y)+tan(19) = 0$$which gives you that $tan(y) = 0.890833$ or $tan(y)= 0.561272$ which means $x = 27times0.890833$ or $x= 27times 0.561272$.
So the question is now which $x$ should I pick and why? and is this more complicated than it should be?
geometry proof-verification proof-explanation triangle alternative-proof
$endgroup$
$begingroup$
Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
$endgroup$
– poetasis
Feb 10 at 16:39
$begingroup$
Why do you assume there is only one solution?
$endgroup$
– fleablood
Feb 10 at 17:46
add a comment |
$begingroup$
I have the following triangle and I'm supposed to find the value of x.
First thought that came to mind is to use the the following tangent equation
$$tan(y)=frac{x}{27}$$ and $$tan(19+y) = frac{2x}{27}$$ which implies that $$tan(19+y) =2tan(y)$$ and solve for $y$ and once I've found $y$ I can easily find $x$.
I don't have an easy solution for the last equation and I feel that I'm complicating things unnecessarily. (It's a grade 10 geometry question).
Edit: Using the fact $tan(a+b) = dfrac{tan(a)+tan(b)}{1-tan(a)tan(b)}$ yields the following quadratic equation $$2tan(19)tan^2(y)-tan(y)+tan(19) = 0$$which gives you that $tan(y) = 0.890833$ or $tan(y)= 0.561272$ which means $x = 27times0.890833$ or $x= 27times 0.561272$.
So the question is now which $x$ should I pick and why? and is this more complicated than it should be?
geometry proof-verification proof-explanation triangle alternative-proof
$endgroup$
I have the following triangle and I'm supposed to find the value of x.
First thought that came to mind is to use the the following tangent equation
$$tan(y)=frac{x}{27}$$ and $$tan(19+y) = frac{2x}{27}$$ which implies that $$tan(19+y) =2tan(y)$$ and solve for $y$ and once I've found $y$ I can easily find $x$.
I don't have an easy solution for the last equation and I feel that I'm complicating things unnecessarily. (It's a grade 10 geometry question).
Edit: Using the fact $tan(a+b) = dfrac{tan(a)+tan(b)}{1-tan(a)tan(b)}$ yields the following quadratic equation $$2tan(19)tan^2(y)-tan(y)+tan(19) = 0$$which gives you that $tan(y) = 0.890833$ or $tan(y)= 0.561272$ which means $x = 27times0.890833$ or $x= 27times 0.561272$.
So the question is now which $x$ should I pick and why? and is this more complicated than it should be?
geometry proof-verification proof-explanation triangle alternative-proof
geometry proof-verification proof-explanation triangle alternative-proof
edited Feb 10 at 16:24
Anirban Niloy
8271218
8271218
asked Feb 10 at 15:24
unknown1unknown1
747
747
$begingroup$
Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
$endgroup$
– poetasis
Feb 10 at 16:39
$begingroup$
Why do you assume there is only one solution?
$endgroup$
– fleablood
Feb 10 at 17:46
add a comment |
$begingroup$
Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
$endgroup$
– poetasis
Feb 10 at 16:39
$begingroup$
Why do you assume there is only one solution?
$endgroup$
– fleablood
Feb 10 at 17:46
$begingroup$
Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
$endgroup$
– poetasis
Feb 10 at 16:39
$begingroup$
Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
$endgroup$
– poetasis
Feb 10 at 16:39
$begingroup$
Why do you assume there is only one solution?
$endgroup$
– fleablood
Feb 10 at 17:46
$begingroup$
Why do you assume there is only one solution?
$endgroup$
– fleablood
Feb 10 at 17:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As you noticed $2tan(y)=tan(19+y)$. Hence
$$2tan(y)=frac{2x}{27}=tan(19+y)=frac{tan(19)+tan(y)}{1-tan(y)tan(19)}=frac{tan(19)+frac{x}{27}}{1-tan(19)·frac{x}{27}}=frac{27tan(19)+x}{27-tan(19)x}$$ $$iff frac{2x}{27}=frac{27tan(19)+x}{27-tan(19)x}iff 54x-2tan(19)·x^2=729tan(19)+27x$$ $$iff 2tan(19)·x^2-27x+729tan(19)=0$$
This gives $$x=15.15...lor x=24.05...$$
Now, as you can see in the following image (made with geogebra), both values are correct:
$endgroup$
add a comment |
$begingroup$
I think you can just use the sinus rule and pythagorean theorem, then:
$$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}}{sin(71-y)}$$
And we see:
$$sin(71-y)=frac{27}{sqrt{4x^2+27^2}}$$
Thus, if we substitute:
$$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}sqrt{4x^2+27^2}}{27}$$
Solving for $x$:
$$x=15.15...lor x=24.05...$$
... witch is in compliance with your answer.
PS.
Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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active
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active
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votes
$begingroup$
As you noticed $2tan(y)=tan(19+y)$. Hence
$$2tan(y)=frac{2x}{27}=tan(19+y)=frac{tan(19)+tan(y)}{1-tan(y)tan(19)}=frac{tan(19)+frac{x}{27}}{1-tan(19)·frac{x}{27}}=frac{27tan(19)+x}{27-tan(19)x}$$ $$iff frac{2x}{27}=frac{27tan(19)+x}{27-tan(19)x}iff 54x-2tan(19)·x^2=729tan(19)+27x$$ $$iff 2tan(19)·x^2-27x+729tan(19)=0$$
This gives $$x=15.15...lor x=24.05...$$
Now, as you can see in the following image (made with geogebra), both values are correct:
$endgroup$
add a comment |
$begingroup$
As you noticed $2tan(y)=tan(19+y)$. Hence
$$2tan(y)=frac{2x}{27}=tan(19+y)=frac{tan(19)+tan(y)}{1-tan(y)tan(19)}=frac{tan(19)+frac{x}{27}}{1-tan(19)·frac{x}{27}}=frac{27tan(19)+x}{27-tan(19)x}$$ $$iff frac{2x}{27}=frac{27tan(19)+x}{27-tan(19)x}iff 54x-2tan(19)·x^2=729tan(19)+27x$$ $$iff 2tan(19)·x^2-27x+729tan(19)=0$$
This gives $$x=15.15...lor x=24.05...$$
Now, as you can see in the following image (made with geogebra), both values are correct:
$endgroup$
add a comment |
$begingroup$
As you noticed $2tan(y)=tan(19+y)$. Hence
$$2tan(y)=frac{2x}{27}=tan(19+y)=frac{tan(19)+tan(y)}{1-tan(y)tan(19)}=frac{tan(19)+frac{x}{27}}{1-tan(19)·frac{x}{27}}=frac{27tan(19)+x}{27-tan(19)x}$$ $$iff frac{2x}{27}=frac{27tan(19)+x}{27-tan(19)x}iff 54x-2tan(19)·x^2=729tan(19)+27x$$ $$iff 2tan(19)·x^2-27x+729tan(19)=0$$
This gives $$x=15.15...lor x=24.05...$$
Now, as you can see in the following image (made with geogebra), both values are correct:
$endgroup$
As you noticed $2tan(y)=tan(19+y)$. Hence
$$2tan(y)=frac{2x}{27}=tan(19+y)=frac{tan(19)+tan(y)}{1-tan(y)tan(19)}=frac{tan(19)+frac{x}{27}}{1-tan(19)·frac{x}{27}}=frac{27tan(19)+x}{27-tan(19)x}$$ $$iff frac{2x}{27}=frac{27tan(19)+x}{27-tan(19)x}iff 54x-2tan(19)·x^2=729tan(19)+27x$$ $$iff 2tan(19)·x^2-27x+729tan(19)=0$$
This gives $$x=15.15...lor x=24.05...$$
Now, as you can see in the following image (made with geogebra), both values are correct:
answered Feb 10 at 17:03
Dr. MathvaDr. Mathva
2,271526
2,271526
add a comment |
add a comment |
$begingroup$
I think you can just use the sinus rule and pythagorean theorem, then:
$$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}}{sin(71-y)}$$
And we see:
$$sin(71-y)=frac{27}{sqrt{4x^2+27^2}}$$
Thus, if we substitute:
$$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}sqrt{4x^2+27^2}}{27}$$
Solving for $x$:
$$x=15.15...lor x=24.05...$$
... witch is in compliance with your answer.
PS.
Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.
$endgroup$
add a comment |
$begingroup$
I think you can just use the sinus rule and pythagorean theorem, then:
$$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}}{sin(71-y)}$$
And we see:
$$sin(71-y)=frac{27}{sqrt{4x^2+27^2}}$$
Thus, if we substitute:
$$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}sqrt{4x^2+27^2}}{27}$$
Solving for $x$:
$$x=15.15...lor x=24.05...$$
... witch is in compliance with your answer.
PS.
Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.
$endgroup$
add a comment |
$begingroup$
I think you can just use the sinus rule and pythagorean theorem, then:
$$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}}{sin(71-y)}$$
And we see:
$$sin(71-y)=frac{27}{sqrt{4x^2+27^2}}$$
Thus, if we substitute:
$$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}sqrt{4x^2+27^2}}{27}$$
Solving for $x$:
$$x=15.15...lor x=24.05...$$
... witch is in compliance with your answer.
PS.
Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.
$endgroup$
I think you can just use the sinus rule and pythagorean theorem, then:
$$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}}{sin(71-y)}$$
And we see:
$$sin(71-y)=frac{27}{sqrt{4x^2+27^2}}$$
Thus, if we substitute:
$$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}sqrt{4x^2+27^2}}{27}$$
Solving for $x$:
$$x=15.15...lor x=24.05...$$
... witch is in compliance with your answer.
PS.
Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.
edited Feb 10 at 17:16
answered Feb 10 at 16:40
MaxMax
564317
564317
add a comment |
add a comment |
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$begingroup$
Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
$endgroup$
– poetasis
Feb 10 at 16:39
$begingroup$
Why do you assume there is only one solution?
$endgroup$
– fleablood
Feb 10 at 17:46