How to use double angle identities to find length 'd'?
$begingroup$
Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.
A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:
Edit: Note that triangle ACB is not a right angled triangle. Should now read:
Note that triangle ACD is not a right angled triangle.
By using $$cos 2x=2cos^2 x-1$$
My working is:
$$cos x=frac d3$$
$$cos 2x=frac d3$$
therefore:
$$frac d3=2(frac d3)^2-1$$
Now becomes solvable as a quadratic equation
trigonometry
asked Dec 11 '18 at 16:08
HawkinsCivilHawkinsCivil
14
$endgroup$
|
show 2 more comments
$begingroup$
Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.
A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:
Edit: Note that triangle ACB is not a right angled triangle. Should now read:
Note that triangle ACD is not a right angled triangle.
By using $$cos 2x=2cos^2 x-1$$
My working is:
$$cos x=frac d3$$
$$cos 2x=frac d3$$
therefore:
$$frac d3=2(frac d3)^2-1$$
Now becomes solvable as a quadratic equation
trigonometry
asked Dec 11 '18 at 16:08
HawkinsCivilHawkinsCivil
14
$endgroup$
$begingroup$
I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
$endgroup$
– hardmath
Dec 11 '18 at 16:39
$begingroup$
Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:40
$begingroup$
This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
$endgroup$
– Love Invariants
Dec 11 '18 at 16:53
$begingroup$
Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:56
$begingroup$
Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
$endgroup$
– Love Invariants
Dec 11 '18 at 17:24
|
show 2 more comments
$begingroup$
Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.
A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:
Edit: Note that triangle ACB is not a right angled triangle. Should now read:
Note that triangle ACD is not a right angled triangle.
By using $$cos 2x=2cos^2 x-1$$
My working is:
$$cos x=frac d3$$
$$cos 2x=frac d3$$
therefore:
$$frac d3=2(frac d3)^2-1$$
Now becomes solvable as a quadratic equation
trigonometry
asked Dec 11 '18 at 16:08
HawkinsCivilHawkinsCivil
14
$endgroup$
Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.
A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:
Edit: Note that triangle ACB is not a right angled triangle. Should now read:
Note that triangle ACD is not a right angled triangle.
By using $$cos 2x=2cos^2 x-1$$
My working is:
$$cos x=frac d3$$
$$cos 2x=frac d3$$
therefore:
$$frac d3=2(frac d3)^2-1$$
Now becomes solvable as a quadratic equation
trigonometry
trigonometry
asked Dec 11 '18 at 16:08
HawkinsCivilHawkinsCivil
14
asked Dec 11 '18 at 16:08
HawkinsCivilHawkinsCivil
14
edited Dec 12 '18 at 22:01
HawkinsCivil
asked Dec 11 '18 at 16:08
HawkinsCivilHawkinsCivil
14
asked Dec 11 '18 at 16:08
HawkinsCivilHawkinsCivil
14
asked Dec 11 '18 at 16:08
HawkinsCivilHawkinsCivil
14
14
$begingroup$
I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
$endgroup$
– hardmath
Dec 11 '18 at 16:39
$begingroup$
Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:40
$begingroup$
This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
$endgroup$
– Love Invariants
Dec 11 '18 at 16:53
$begingroup$
Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:56
$begingroup$
Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
$endgroup$
– Love Invariants
Dec 11 '18 at 17:24
|
show 2 more comments
$begingroup$
I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
$endgroup$
– hardmath
Dec 11 '18 at 16:39
$begingroup$
Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:40
$begingroup$
This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
$endgroup$
– Love Invariants
Dec 11 '18 at 16:53
$begingroup$
Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:56
$begingroup$
Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
$endgroup$
– Love Invariants
Dec 11 '18 at 17:24
$begingroup$
I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
$endgroup$
– hardmath
Dec 11 '18 at 16:39
$begingroup$
I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
$endgroup$
– hardmath
Dec 11 '18 at 16:39
$begingroup$
Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:40
$begingroup$
Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:40
$begingroup$
This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
$endgroup$
– Love Invariants
Dec 11 '18 at 16:53
$begingroup$
This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
$endgroup$
– Love Invariants
Dec 11 '18 at 16:53
$begingroup$
Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:56
$begingroup$
Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:56
$begingroup$
Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
$endgroup$
– Love Invariants
Dec 11 '18 at 17:24
$begingroup$
Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
$endgroup$
– Love Invariants
Dec 11 '18 at 17:24
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"
Using the law of cosines in triangles $ABC$ and $ACD$ we get
$$c^2=2^2+3^2-2times 2times 3 cos x$$
$$b^2=d^2+2^2-4dcos x$$
so that $$cos x=dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-dfrac{4d(13-c^2)}{12}$$
The last equation is quadratic in $d$ and easy to solve.
answered Dec 11 '18 at 17:29
BPPBPP
2,159927
$endgroup$
$begingroup$
Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
$endgroup$
– BPP
Dec 11 '18 at 20:09
add a comment |
$begingroup$
To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:
$$x = frac{-b pm sqrt{b^2 - 4ac} }{2a}$$
answered Dec 20 '18 at 10:57
HawkinsCivilHawkinsCivil
14
$endgroup$
add a comment |
$begingroup$
$cos(x)neq frac{2}{3}$ because the angle $widehat{ACD} neq frac{pi}{2}$.
edited Dec 11 '18 at 17:29
Davide Giraudo
126k16150261
answered Dec 11 '18 at 16:49
M.gaussM.gauss
91
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"
Using the law of cosines in triangles $ABC$ and $ACD$ we get
$$c^2=2^2+3^2-2times 2times 3 cos x$$
$$b^2=d^2+2^2-4dcos x$$
so that $$cos x=dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-dfrac{4d(13-c^2)}{12}$$
The last equation is quadratic in $d$ and easy to solve.
answered Dec 11 '18 at 17:29
BPPBPP
2,159927
$endgroup$
$begingroup$
Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
$endgroup$
– BPP
Dec 11 '18 at 20:09
add a comment |
$begingroup$
I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"
Using the law of cosines in triangles $ABC$ and $ACD$ we get
$$c^2=2^2+3^2-2times 2times 3 cos x$$
$$b^2=d^2+2^2-4dcos x$$
so that $$cos x=dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-dfrac{4d(13-c^2)}{12}$$
The last equation is quadratic in $d$ and easy to solve.
answered Dec 11 '18 at 17:29
BPPBPP
2,159927
$endgroup$
$begingroup$
Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
$endgroup$
– BPP
Dec 11 '18 at 20:09
add a comment |
$begingroup$
I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"
Using the law of cosines in triangles $ABC$ and $ACD$ we get
$$c^2=2^2+3^2-2times 2times 3 cos x$$
$$b^2=d^2+2^2-4dcos x$$
so that $$cos x=dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-dfrac{4d(13-c^2)}{12}$$
The last equation is quadratic in $d$ and easy to solve.
answered Dec 11 '18 at 17:29
BPPBPP
2,159927
$endgroup$
I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"
Using the law of cosines in triangles $ABC$ and $ACD$ we get
$$c^2=2^2+3^2-2times 2times 3 cos x$$
$$b^2=d^2+2^2-4dcos x$$
so that $$cos x=dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-dfrac{4d(13-c^2)}{12}$$
The last equation is quadratic in $d$ and easy to solve.
answered Dec 11 '18 at 17:29
BPPBPP
2,159927
answered Dec 11 '18 at 17:29
BPPBPP
2,159927
answered Dec 11 '18 at 17:29
BPPBPP
2,159927
answered Dec 11 '18 at 17:29
BPPBPP
2,159927
2,159927
$begingroup$
Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
$endgroup$
– BPP
Dec 11 '18 at 20:09
add a comment |
$begingroup$
Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
$endgroup$
– BPP
Dec 11 '18 at 20:09
$begingroup$
Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
$endgroup$
– BPP
Dec 11 '18 at 20:09
$begingroup$
Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
$endgroup$
– BPP
Dec 11 '18 at 20:09
add a comment |
$begingroup$
To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:
$$x = frac{-b pm sqrt{b^2 - 4ac} }{2a}$$
answered Dec 20 '18 at 10:57
HawkinsCivilHawkinsCivil
14
$endgroup$
add a comment |
$begingroup$
To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:
$$x = frac{-b pm sqrt{b^2 - 4ac} }{2a}$$
answered Dec 20 '18 at 10:57
HawkinsCivilHawkinsCivil
14
$endgroup$
add a comment |
$begingroup$
To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:
$$x = frac{-b pm sqrt{b^2 - 4ac} }{2a}$$
answered Dec 20 '18 at 10:57
HawkinsCivilHawkinsCivil
14
$endgroup$
To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:
$$x = frac{-b pm sqrt{b^2 - 4ac} }{2a}$$
answered Dec 20 '18 at 10:57
HawkinsCivilHawkinsCivil
14
answered Dec 20 '18 at 10:57
HawkinsCivilHawkinsCivil
14
answered Dec 20 '18 at 10:57
HawkinsCivilHawkinsCivil
14
answered Dec 20 '18 at 10:57
HawkinsCivilHawkinsCivil
14
14
add a comment |
add a comment |
$begingroup$
$cos(x)neq frac{2}{3}$ because the angle $widehat{ACD} neq frac{pi}{2}$.
edited Dec 11 '18 at 17:29
Davide Giraudo
126k16150261
answered Dec 11 '18 at 16:49
M.gaussM.gauss
91
$endgroup$
add a comment |
$begingroup$
$cos(x)neq frac{2}{3}$ because the angle $widehat{ACD} neq frac{pi}{2}$.
edited Dec 11 '18 at 17:29
Davide Giraudo
126k16150261
answered Dec 11 '18 at 16:49
M.gaussM.gauss
91
$endgroup$
add a comment |
$begingroup$
$cos(x)neq frac{2}{3}$ because the angle $widehat{ACD} neq frac{pi}{2}$.
edited Dec 11 '18 at 17:29
Davide Giraudo
126k16150261
answered Dec 11 '18 at 16:49
M.gaussM.gauss
91
$endgroup$
$cos(x)neq frac{2}{3}$ because the angle $widehat{ACD} neq frac{pi}{2}$.
edited Dec 11 '18 at 17:29
Davide Giraudo
126k16150261
answered Dec 11 '18 at 16:49
M.gaussM.gauss
91
edited Dec 11 '18 at 17:29
Davide Giraudo
126k16150261
edited Dec 11 '18 at 17:29
Davide Giraudo
126k16150261
edited Dec 11 '18 at 17:29
Davide Giraudo
126k16150261
126k16150261
answered Dec 11 '18 at 16:49
M.gaussM.gauss
91
answered Dec 11 '18 at 16:49
M.gaussM.gauss
91
answered Dec 11 '18 at 16:49
M.gaussM.gauss
91
91
add a comment |
add a comment |
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$begingroup$
I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
$endgroup$
– hardmath
Dec 11 '18 at 16:39
$begingroup$
Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:40
$begingroup$
This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
$endgroup$
– Love Invariants
Dec 11 '18 at 16:53
$begingroup$
Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:56
$begingroup$
Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
$endgroup$
– Love Invariants
Dec 11 '18 at 17:24