Mixing
Solve for the angle x?
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The problem seems really impossible to me, despite solving many problems on triangles. So I hope that you can solve it.
src:https://www.instagram.com/p/BtUNj3IBIW5/
geometry triangles angle
edited Feb 3 at 22:14
user376343
3,9834829
asked Feb 3 at 12:54
Khaled OqabKhaled Oqab
515
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add a comment |
$begingroup$
The problem seems really impossible to me, despite solving many problems on triangles. So I hope that you can solve it.
src:https://www.instagram.com/p/BtUNj3IBIW5/
geometry triangles angle
edited Feb 3 at 22:14
user376343
3,9834829
asked Feb 3 at 12:54
Khaled OqabKhaled Oqab
515
$endgroup$
$begingroup$
I got $$x=frac{pi}{3}$$, i this right?
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– Dr. Sonnhard Graubner
Feb 3 at 13:19
$begingroup$
actually you will find that you can scale the diagram and that the answer will be x=40 degrees, but what my question really is " how to find it mathematically"
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– Khaled Oqab
Feb 3 at 13:42
$begingroup$
Ok, i will check my answer, thank you for the information!
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 13:49
add a comment |
$begingroup$
The problem seems really impossible to me, despite solving many problems on triangles. So I hope that you can solve it.
src:https://www.instagram.com/p/BtUNj3IBIW5/
geometry triangles angle
edited Feb 3 at 22:14
user376343
3,9834829
asked Feb 3 at 12:54
Khaled OqabKhaled Oqab
515
$endgroup$
The problem seems really impossible to me, despite solving many problems on triangles. So I hope that you can solve it.
src:https://www.instagram.com/p/BtUNj3IBIW5/
geometry triangles angle
geometry triangles angle
edited Feb 3 at 22:14
user376343
3,9834829
asked Feb 3 at 12:54
Khaled OqabKhaled Oqab
515
edited Feb 3 at 22:14
user376343
3,9834829
asked Feb 3 at 12:54
Khaled OqabKhaled Oqab
515
edited Feb 3 at 22:14
user376343
3,9834829
edited Feb 3 at 22:14
user376343
3,9834829
edited Feb 3 at 22:14
user376343
3,9834829
3,9834829
asked Feb 3 at 12:54
Khaled OqabKhaled Oqab
515
asked Feb 3 at 12:54
Khaled OqabKhaled Oqab
515
asked Feb 3 at 12:54
Khaled OqabKhaled Oqab
515
515
$begingroup$
I got $$x=frac{pi}{3}$$, i this right?
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 13:19
$begingroup$
actually you will find that you can scale the diagram and that the answer will be x=40 degrees, but what my question really is " how to find it mathematically"
$endgroup$
– Khaled Oqab
Feb 3 at 13:42
$begingroup$
Ok, i will check my answer, thank you for the information!
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 13:49
add a comment |
$begingroup$
I got $$x=frac{pi}{3}$$, i this right?
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 13:19
$begingroup$
actually you will find that you can scale the diagram and that the answer will be x=40 degrees, but what my question really is " how to find it mathematically"
$endgroup$
– Khaled Oqab
Feb 3 at 13:42
$begingroup$
Ok, i will check my answer, thank you for the information!
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 13:49
$begingroup$
I got $$x=frac{pi}{3}$$, i this right?
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 13:19
$begingroup$
I got $$x=frac{pi}{3}$$, i this right?
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 13:19
$begingroup$
actually you will find that you can scale the diagram and that the answer will be x=40 degrees, but what my question really is " how to find it mathematically"
$endgroup$
– Khaled Oqab
Feb 3 at 13:42
$begingroup$
actually you will find that you can scale the diagram and that the answer will be x=40 degrees, but what my question really is " how to find it mathematically"
$endgroup$
– Khaled Oqab
Feb 3 at 13:42
$begingroup$
Ok, i will check my answer, thank you for the information!
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 13:49
$begingroup$
Ok, i will check my answer, thank you for the information!
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 13:49
add a comment |
3 Answers
3
active
oldest
votes
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Unfortunately, some case analyses and contradiction things must be done and I wasn't able to come up with any nice and clean way.
First create triangle $ACG$ congruent to $ABC$ and label $F$ as shown in the picture. Mirror $D$ along axis $AC$ to point $P$. Connect necessary points as shown in the picture.
In trapezoid $ADCG$, triangles $ACG$ and $DGC$ are symmetric so congruent as well. So $DGC$ is also a triangle congruent to $ABC$ where angle $angle DGC=x$
At the same time, $PA = AD = EC, AG = CD, angle PAG = 90 - {xover 2} - x = angle ECD implies triangle PAG cong triangle ECD$.
Furthermore it's easy to see $DE=PG$ and $triangle DEG cong triangle GPC$.
Now let $angle GDE = angle CGP = a$.
With some angle tracing (with the known congruences every single angle can be expressed in terms of x), $a = 40 - {xover 2}$, and $angle DEP = 280 - 3x$.
Case $1$: $EP$ is not parallel to $DG$ and intersect $DG$ on the extension ray from $D$ to $G$. Then this means $angle GDE + angle DEP < 180$ which means $320 - {7xover 2} < 180$ so $x > 40$ and $a < 20$.
Therefore $angle DGP = x - a > 20 > a = angle GDE$. However since $DE = PG$ and $angle DGP > angle GDE$, $P$ would have a further distance to $DG$ and $E$ and $PE$ would intersect $GD$ on the extension ray from $G$ to $D$. contradiction.
Case $2$: $EP$ is not parallel to $DG$ and intersect $GD$ on the extension ray from $G$ to $D$. Then this means $angle GDE + angle DEP > 180$ which means $320 - {7xover 2} > 180$ so $x < 40$ and $a > 20$. All similar thing and we can conclude $angle GDE > angle DGP$ so $EP$ must intersect $DG$ on the extension from $D$ to $G$. contradiction.
So $EP$ must be parallel to $DG$ and $320 - {7xover 2} = 180$ and $x=40$.
answered Feb 4 at 16:03
cr001cr001
7,871517
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add a comment |
$begingroup$
With the notations on the figure :
$X=widehat{BAC}$
$beta=widehat{CBA}=frac{pi-X}{2}$
$gamma=widehat{DCB}=pi-2beta=X$
$delta=widehat{ACD}=beta-gamma=frac{pi-3X}{2}$
Without lost of generality, let $BC=1$
$BD=AE=2cos(beta)=2sin(frac{X}{2})$
$BA=frac{1}{2cos(beta)}=frac{1}{2sin(frac{X}{2})}$
$CE=DA=BA-BD=frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})$
$EH=CEsin(delta)= (frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})) cos(frac{3X}{2})$
$CH=CEcos(delta)= (frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2}))sin(frac{3X}{2})$
$HD=CD-CH=1-frac{sin(frac{3X}{2} )}{2sin(frac{X}{2})}+2sin(frac{X}{2})sin(frac{3X}{2})$
$tan(Y)=tan(widehat{CDE})=frac{EH}{HD}=frac{(frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})) cos(frac{3X}{2})}{ 1-frac{sin(frac{3X}{2} )}{2sin(frac{X}{2})}+2sin(frac{X}{2})sin(frac{3X}{2}) }$
The relationship between angles $X$ and $Y$ is :
$$tan(Y)=frac{(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})}{2sin(frac{X}{2}) -sin(frac{3X}{2} )+4sin^2(frac{X}{2})sin(frac{3X}{2}) } $$
$$cot(Y)=frac{2sin(frac{X}{2}) } {(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})} -tan(frac{3X}{2}) tag 1$$
The function $Y(X)$ is represented on the next figure.
For example, if we set $X=pifrac{40}{180}$ that is an angle of $40°$ , the numerical result is :
$$Ysimeq 0.872664625997165quadtoquad 180frac{Y}{pi}simeq 50.00000000000003$$
Thus, from numerical calculus, at $Y=50°$ corresponds $X=40°$ with a very good accuracy.
Analytically one have to prove that the equation $(1)$ returns exactly $Y=pifrac{50}{180}$ for $X=pifrac{40}{180}$ .
Consider la function :
$$f(X,Y)=frac{2sin(frac{X}{2}) } {(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})} -tan(frac{3X}{2}) -cot(Y) $$
we have to prove that this function is $=0$ when $Y=pifrac{50}{180}$ and $X=pifrac{40}{180}$
$$f(frac{2pi}{9},frac{5pi}{18}) =frac{2sin(frac{pi}{9}) } {(1-4sin^2(frac{pi}{9})) cos(frac{pi}{3})} -tan(frac{pi}{3}) -cot(frac{5pi}{18}) $$
$$f((frac{2pi}{9},frac{5pi}{18}) =frac{4sin(frac{pi}{9}) } {(1-4sin^2(frac{pi}{9}))} -sqrt{3} -cot(frac{5pi}{18}) $$
Let $s=sin(frac{pi}{18})$
$sin(frac{pi}{9})=2ssqrt{1-s^2}$
$cos(frac{5pi}{18})= (16s^4-12s^2+1)sqrt{1-s^2}$
$sin(frac{5pi}{18})=(16s^5-20s^3+5s)$
$$f(frac{2pi}{9},frac{5pi}{18}) =frac{8ssqrt{1-s^2} } {1-4(2ssqrt{1-s^2})^2} -sqrt{3} -frac {(16s^4-12s^2+1)sqrt{1-s^2}}{16s^5-20s^3+5s} $$
In fact, we have to check if one of the roots of the equation :
$$frac{8ssqrt{1-s^2} } {1-4(2ssqrt{1-s^2})^2} -sqrt{3} -frac {(16s^4-12s^2+1)sqrt{1-s^2}}{16s^5-20s^3+5s}=0 tag 2$$
is equal to $sin(frac{pi}{18})$ .
Expanding and factoring transform the equation to :
$$(8s^3-6s-1)(8s^3-6s+1)(4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1)=0$$
For $s=sin(frac{pi}{18})$ the polynomials $8s^3-6s+1neq 0$ and $4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1neq 0$ .
Only the equation :
$$8s^3-6s+1=0$$
is likely to have a root around to $sin(frac{pi}{18})$. Then we have to prove that it is the case.
We put into the equation $s=sin(frac{pi}{18})=frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i}$
$$8s^3-6s+1=8left(frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i} right)^3-6left(frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i} right)+1$$
After expansion and simplification all terms cancel. We get : $8s^3-6s+1=0$ .
Thus $s=sin(frac{pi}{18})$ is a root of equation $(2)$ and as a consequence $f(frac{2pi}{9},frac{5pi}{18}) =0$ ,
This is the analytical proof that $X=40°$ and $Y=50°$ is an exact solution.
answered Feb 4 at 10:35
JJacquelinJJacquelin
45.7k21858
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Here is a synthetic solution.
Let $angle BAC = x$ and $angle DCE = y$
Now, $angle ABC = 60 + frac{y}{3}$. If we prove $y= 30$, the proof would be complete. Let $BC = a$
Note that $$BD = 2a cos ({60 +frac{y}{3}})$$
and since $AB= frac {a}{2cos ({60 + frac{y}{3}})}$, we can calculate after simplifying and using the property
:$frac {4cos^2 t -1}{2cos t} = frac {3- 4sin^2 t}{2cos t}= frac {sin 3t}{sin 2t}$
$$frac {AD}{a} = frac {sin y}{cos ({30 + frac {2y}{3}})}$$
By sine rule in $Delta DEC$, and using the above result,
$$sin{y}sin({y+50})= sin{50}cos({30+frac{2y}{3}})$$
And so,
$$2sin{y}sin({y+50})= 2sin{50}cos({30+frac{2y}{3}})$$
if $y=30$, then this equation is satisfied because RHS becomes $sin 100$ and LHS becomes $sin 80$ and they are obviously equal.
Therefore, $x=40$
Note:
To see that that equation has no other solutions than $y=30$, I suggest using Wolfram Alpha.
answered Feb 5 at 11:32
DhvanitDhvanit
10010
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3 Answers
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3 Answers
3
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$begingroup$
Unfortunately, some case analyses and contradiction things must be done and I wasn't able to come up with any nice and clean way.
First create triangle $ACG$ congruent to $ABC$ and label $F$ as shown in the picture. Mirror $D$ along axis $AC$ to point $P$. Connect necessary points as shown in the picture.
In trapezoid $ADCG$, triangles $ACG$ and $DGC$ are symmetric so congruent as well. So $DGC$ is also a triangle congruent to $ABC$ where angle $angle DGC=x$
At the same time, $PA = AD = EC, AG = CD, angle PAG = 90 - {xover 2} - x = angle ECD implies triangle PAG cong triangle ECD$.
Furthermore it's easy to see $DE=PG$ and $triangle DEG cong triangle GPC$.
Now let $angle GDE = angle CGP = a$.
With some angle tracing (with the known congruences every single angle can be expressed in terms of x), $a = 40 - {xover 2}$, and $angle DEP = 280 - 3x$.
Case $1$: $EP$ is not parallel to $DG$ and intersect $DG$ on the extension ray from $D$ to $G$. Then this means $angle GDE + angle DEP < 180$ which means $320 - {7xover 2} < 180$ so $x > 40$ and $a < 20$.
Therefore $angle DGP = x - a > 20 > a = angle GDE$. However since $DE = PG$ and $angle DGP > angle GDE$, $P$ would have a further distance to $DG$ and $E$ and $PE$ would intersect $GD$ on the extension ray from $G$ to $D$. contradiction.
Case $2$: $EP$ is not parallel to $DG$ and intersect $GD$ on the extension ray from $G$ to $D$. Then this means $angle GDE + angle DEP > 180$ which means $320 - {7xover 2} > 180$ so $x < 40$ and $a > 20$. All similar thing and we can conclude $angle GDE > angle DGP$ so $EP$ must intersect $DG$ on the extension from $D$ to $G$. contradiction.
So $EP$ must be parallel to $DG$ and $320 - {7xover 2} = 180$ and $x=40$.
answered Feb 4 at 16:03
cr001cr001
7,871517
$endgroup$
add a comment |
$begingroup$
Unfortunately, some case analyses and contradiction things must be done and I wasn't able to come up with any nice and clean way.
First create triangle $ACG$ congruent to $ABC$ and label $F$ as shown in the picture. Mirror $D$ along axis $AC$ to point $P$. Connect necessary points as shown in the picture.
In trapezoid $ADCG$, triangles $ACG$ and $DGC$ are symmetric so congruent as well. So $DGC$ is also a triangle congruent to $ABC$ where angle $angle DGC=x$
At the same time, $PA = AD = EC, AG = CD, angle PAG = 90 - {xover 2} - x = angle ECD implies triangle PAG cong triangle ECD$.
Furthermore it's easy to see $DE=PG$ and $triangle DEG cong triangle GPC$.
Now let $angle GDE = angle CGP = a$.
With some angle tracing (with the known congruences every single angle can be expressed in terms of x), $a = 40 - {xover 2}$, and $angle DEP = 280 - 3x$.
Case $1$: $EP$ is not parallel to $DG$ and intersect $DG$ on the extension ray from $D$ to $G$. Then this means $angle GDE + angle DEP < 180$ which means $320 - {7xover 2} < 180$ so $x > 40$ and $a < 20$.
Therefore $angle DGP = x - a > 20 > a = angle GDE$. However since $DE = PG$ and $angle DGP > angle GDE$, $P$ would have a further distance to $DG$ and $E$ and $PE$ would intersect $GD$ on the extension ray from $G$ to $D$. contradiction.
Case $2$: $EP$ is not parallel to $DG$ and intersect $GD$ on the extension ray from $G$ to $D$. Then this means $angle GDE + angle DEP > 180$ which means $320 - {7xover 2} > 180$ so $x < 40$ and $a > 20$. All similar thing and we can conclude $angle GDE > angle DGP$ so $EP$ must intersect $DG$ on the extension from $D$ to $G$. contradiction.
So $EP$ must be parallel to $DG$ and $320 - {7xover 2} = 180$ and $x=40$.
answered Feb 4 at 16:03
cr001cr001
7,871517
$endgroup$
add a comment |
$begingroup$
Unfortunately, some case analyses and contradiction things must be done and I wasn't able to come up with any nice and clean way.
First create triangle $ACG$ congruent to $ABC$ and label $F$ as shown in the picture. Mirror $D$ along axis $AC$ to point $P$. Connect necessary points as shown in the picture.
In trapezoid $ADCG$, triangles $ACG$ and $DGC$ are symmetric so congruent as well. So $DGC$ is also a triangle congruent to $ABC$ where angle $angle DGC=x$
At the same time, $PA = AD = EC, AG = CD, angle PAG = 90 - {xover 2} - x = angle ECD implies triangle PAG cong triangle ECD$.
Furthermore it's easy to see $DE=PG$ and $triangle DEG cong triangle GPC$.
Now let $angle GDE = angle CGP = a$.
With some angle tracing (with the known congruences every single angle can be expressed in terms of x), $a = 40 - {xover 2}$, and $angle DEP = 280 - 3x$.
Case $1$: $EP$ is not parallel to $DG$ and intersect $DG$ on the extension ray from $D$ to $G$. Then this means $angle GDE + angle DEP < 180$ which means $320 - {7xover 2} < 180$ so $x > 40$ and $a < 20$.
Therefore $angle DGP = x - a > 20 > a = angle GDE$. However since $DE = PG$ and $angle DGP > angle GDE$, $P$ would have a further distance to $DG$ and $E$ and $PE$ would intersect $GD$ on the extension ray from $G$ to $D$. contradiction.
Case $2$: $EP$ is not parallel to $DG$ and intersect $GD$ on the extension ray from $G$ to $D$. Then this means $angle GDE + angle DEP > 180$ which means $320 - {7xover 2} > 180$ so $x < 40$ and $a > 20$. All similar thing and we can conclude $angle GDE > angle DGP$ so $EP$ must intersect $DG$ on the extension from $D$ to $G$. contradiction.
So $EP$ must be parallel to $DG$ and $320 - {7xover 2} = 180$ and $x=40$.
answered Feb 4 at 16:03
cr001cr001
7,871517
$endgroup$
Unfortunately, some case analyses and contradiction things must be done and I wasn't able to come up with any nice and clean way.
First create triangle $ACG$ congruent to $ABC$ and label $F$ as shown in the picture. Mirror $D$ along axis $AC$ to point $P$. Connect necessary points as shown in the picture.
In trapezoid $ADCG$, triangles $ACG$ and $DGC$ are symmetric so congruent as well. So $DGC$ is also a triangle congruent to $ABC$ where angle $angle DGC=x$
At the same time, $PA = AD = EC, AG = CD, angle PAG = 90 - {xover 2} - x = angle ECD implies triangle PAG cong triangle ECD$.
Furthermore it's easy to see $DE=PG$ and $triangle DEG cong triangle GPC$.
Now let $angle GDE = angle CGP = a$.
With some angle tracing (with the known congruences every single angle can be expressed in terms of x), $a = 40 - {xover 2}$, and $angle DEP = 280 - 3x$.
Case $1$: $EP$ is not parallel to $DG$ and intersect $DG$ on the extension ray from $D$ to $G$. Then this means $angle GDE + angle DEP < 180$ which means $320 - {7xover 2} < 180$ so $x > 40$ and $a < 20$.
Therefore $angle DGP = x - a > 20 > a = angle GDE$. However since $DE = PG$ and $angle DGP > angle GDE$, $P$ would have a further distance to $DG$ and $E$ and $PE$ would intersect $GD$ on the extension ray from $G$ to $D$. contradiction.
Case $2$: $EP$ is not parallel to $DG$ and intersect $GD$ on the extension ray from $G$ to $D$. Then this means $angle GDE + angle DEP > 180$ which means $320 - {7xover 2} > 180$ so $x < 40$ and $a > 20$. All similar thing and we can conclude $angle GDE > angle DGP$ so $EP$ must intersect $DG$ on the extension from $D$ to $G$. contradiction.
So $EP$ must be parallel to $DG$ and $320 - {7xover 2} = 180$ and $x=40$.
answered Feb 4 at 16:03
cr001cr001
7,871517
edited Feb 4 at 16:11
answered Feb 4 at 16:03
cr001cr001
7,871517
answered Feb 4 at 16:03
cr001cr001
7,871517
answered Feb 4 at 16:03
cr001cr001
7,871517
7,871517
add a comment |
add a comment |
$begingroup$
With the notations on the figure :
$X=widehat{BAC}$
$beta=widehat{CBA}=frac{pi-X}{2}$
$gamma=widehat{DCB}=pi-2beta=X$
$delta=widehat{ACD}=beta-gamma=frac{pi-3X}{2}$
Without lost of generality, let $BC=1$
$BD=AE=2cos(beta)=2sin(frac{X}{2})$
$BA=frac{1}{2cos(beta)}=frac{1}{2sin(frac{X}{2})}$
$CE=DA=BA-BD=frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})$
$EH=CEsin(delta)= (frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})) cos(frac{3X}{2})$
$CH=CEcos(delta)= (frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2}))sin(frac{3X}{2})$
$HD=CD-CH=1-frac{sin(frac{3X}{2} )}{2sin(frac{X}{2})}+2sin(frac{X}{2})sin(frac{3X}{2})$
$tan(Y)=tan(widehat{CDE})=frac{EH}{HD}=frac{(frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})) cos(frac{3X}{2})}{ 1-frac{sin(frac{3X}{2} )}{2sin(frac{X}{2})}+2sin(frac{X}{2})sin(frac{3X}{2}) }$
The relationship between angles $X$ and $Y$ is :
$$tan(Y)=frac{(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})}{2sin(frac{X}{2}) -sin(frac{3X}{2} )+4sin^2(frac{X}{2})sin(frac{3X}{2}) } $$
$$cot(Y)=frac{2sin(frac{X}{2}) } {(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})} -tan(frac{3X}{2}) tag 1$$
The function $Y(X)$ is represented on the next figure.
For example, if we set $X=pifrac{40}{180}$ that is an angle of $40°$ , the numerical result is :
$$Ysimeq 0.872664625997165quadtoquad 180frac{Y}{pi}simeq 50.00000000000003$$
Thus, from numerical calculus, at $Y=50°$ corresponds $X=40°$ with a very good accuracy.
Analytically one have to prove that the equation $(1)$ returns exactly $Y=pifrac{50}{180}$ for $X=pifrac{40}{180}$ .
Consider la function :
$$f(X,Y)=frac{2sin(frac{X}{2}) } {(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})} -tan(frac{3X}{2}) -cot(Y) $$
we have to prove that this function is $=0$ when $Y=pifrac{50}{180}$ and $X=pifrac{40}{180}$
$$f(frac{2pi}{9},frac{5pi}{18}) =frac{2sin(frac{pi}{9}) } {(1-4sin^2(frac{pi}{9})) cos(frac{pi}{3})} -tan(frac{pi}{3}) -cot(frac{5pi}{18}) $$
$$f((frac{2pi}{9},frac{5pi}{18}) =frac{4sin(frac{pi}{9}) } {(1-4sin^2(frac{pi}{9}))} -sqrt{3} -cot(frac{5pi}{18}) $$
Let $s=sin(frac{pi}{18})$
$sin(frac{pi}{9})=2ssqrt{1-s^2}$
$cos(frac{5pi}{18})= (16s^4-12s^2+1)sqrt{1-s^2}$
$sin(frac{5pi}{18})=(16s^5-20s^3+5s)$
$$f(frac{2pi}{9},frac{5pi}{18}) =frac{8ssqrt{1-s^2} } {1-4(2ssqrt{1-s^2})^2} -sqrt{3} -frac {(16s^4-12s^2+1)sqrt{1-s^2}}{16s^5-20s^3+5s} $$
In fact, we have to check if one of the roots of the equation :
$$frac{8ssqrt{1-s^2} } {1-4(2ssqrt{1-s^2})^2} -sqrt{3} -frac {(16s^4-12s^2+1)sqrt{1-s^2}}{16s^5-20s^3+5s}=0 tag 2$$
is equal to $sin(frac{pi}{18})$ .
Expanding and factoring transform the equation to :
$$(8s^3-6s-1)(8s^3-6s+1)(4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1)=0$$
For $s=sin(frac{pi}{18})$ the polynomials $8s^3-6s+1neq 0$ and $4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1neq 0$ .
Only the equation :
$$8s^3-6s+1=0$$
is likely to have a root around to $sin(frac{pi}{18})$. Then we have to prove that it is the case.
We put into the equation $s=sin(frac{pi}{18})=frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i}$
$$8s^3-6s+1=8left(frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i} right)^3-6left(frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i} right)+1$$
After expansion and simplification all terms cancel. We get : $8s^3-6s+1=0$ .
Thus $s=sin(frac{pi}{18})$ is a root of equation $(2)$ and as a consequence $f(frac{2pi}{9},frac{5pi}{18}) =0$ ,
This is the analytical proof that $X=40°$ and $Y=50°$ is an exact solution.
answered Feb 4 at 10:35
JJacquelinJJacquelin
45.7k21858
$endgroup$
add a comment |
$begingroup$
With the notations on the figure :
$X=widehat{BAC}$
$beta=widehat{CBA}=frac{pi-X}{2}$
$gamma=widehat{DCB}=pi-2beta=X$
$delta=widehat{ACD}=beta-gamma=frac{pi-3X}{2}$
Without lost of generality, let $BC=1$
$BD=AE=2cos(beta)=2sin(frac{X}{2})$
$BA=frac{1}{2cos(beta)}=frac{1}{2sin(frac{X}{2})}$
$CE=DA=BA-BD=frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})$
$EH=CEsin(delta)= (frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})) cos(frac{3X}{2})$
$CH=CEcos(delta)= (frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2}))sin(frac{3X}{2})$
$HD=CD-CH=1-frac{sin(frac{3X}{2} )}{2sin(frac{X}{2})}+2sin(frac{X}{2})sin(frac{3X}{2})$
$tan(Y)=tan(widehat{CDE})=frac{EH}{HD}=frac{(frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})) cos(frac{3X}{2})}{ 1-frac{sin(frac{3X}{2} )}{2sin(frac{X}{2})}+2sin(frac{X}{2})sin(frac{3X}{2}) }$
The relationship between angles $X$ and $Y$ is :
$$tan(Y)=frac{(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})}{2sin(frac{X}{2}) -sin(frac{3X}{2} )+4sin^2(frac{X}{2})sin(frac{3X}{2}) } $$
$$cot(Y)=frac{2sin(frac{X}{2}) } {(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})} -tan(frac{3X}{2}) tag 1$$
The function $Y(X)$ is represented on the next figure.
For example, if we set $X=pifrac{40}{180}$ that is an angle of $40°$ , the numerical result is :
$$Ysimeq 0.872664625997165quadtoquad 180frac{Y}{pi}simeq 50.00000000000003$$
Thus, from numerical calculus, at $Y=50°$ corresponds $X=40°$ with a very good accuracy.
Analytically one have to prove that the equation $(1)$ returns exactly $Y=pifrac{50}{180}$ for $X=pifrac{40}{180}$ .
Consider la function :
$$f(X,Y)=frac{2sin(frac{X}{2}) } {(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})} -tan(frac{3X}{2}) -cot(Y) $$
we have to prove that this function is $=0$ when $Y=pifrac{50}{180}$ and $X=pifrac{40}{180}$
$$f(frac{2pi}{9},frac{5pi}{18}) =frac{2sin(frac{pi}{9}) } {(1-4sin^2(frac{pi}{9})) cos(frac{pi}{3})} -tan(frac{pi}{3}) -cot(frac{5pi}{18}) $$
$$f((frac{2pi}{9},frac{5pi}{18}) =frac{4sin(frac{pi}{9}) } {(1-4sin^2(frac{pi}{9}))} -sqrt{3} -cot(frac{5pi}{18}) $$
Let $s=sin(frac{pi}{18})$
$sin(frac{pi}{9})=2ssqrt{1-s^2}$
$cos(frac{5pi}{18})= (16s^4-12s^2+1)sqrt{1-s^2}$
$sin(frac{5pi}{18})=(16s^5-20s^3+5s)$
$$f(frac{2pi}{9},frac{5pi}{18}) =frac{8ssqrt{1-s^2} } {1-4(2ssqrt{1-s^2})^2} -sqrt{3} -frac {(16s^4-12s^2+1)sqrt{1-s^2}}{16s^5-20s^3+5s} $$
In fact, we have to check if one of the roots of the equation :
$$frac{8ssqrt{1-s^2} } {1-4(2ssqrt{1-s^2})^2} -sqrt{3} -frac {(16s^4-12s^2+1)sqrt{1-s^2}}{16s^5-20s^3+5s}=0 tag 2$$
is equal to $sin(frac{pi}{18})$ .
Expanding and factoring transform the equation to :
$$(8s^3-6s-1)(8s^3-6s+1)(4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1)=0$$
For $s=sin(frac{pi}{18})$ the polynomials $8s^3-6s+1neq 0$ and $4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1neq 0$ .
Only the equation :
$$8s^3-6s+1=0$$
is likely to have a root around to $sin(frac{pi}{18})$. Then we have to prove that it is the case.
We put into the equation $s=sin(frac{pi}{18})=frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i}$
$$8s^3-6s+1=8left(frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i} right)^3-6left(frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i} right)+1$$
After expansion and simplification all terms cancel. We get : $8s^3-6s+1=0$ .
Thus $s=sin(frac{pi}{18})$ is a root of equation $(2)$ and as a consequence $f(frac{2pi}{9},frac{5pi}{18}) =0$ ,
This is the analytical proof that $X=40°$ and $Y=50°$ is an exact solution.
answered Feb 4 at 10:35
JJacquelinJJacquelin
45.7k21858
$endgroup$
add a comment |
$begingroup$
With the notations on the figure :
$X=widehat{BAC}$
$beta=widehat{CBA}=frac{pi-X}{2}$
$gamma=widehat{DCB}=pi-2beta=X$
$delta=widehat{ACD}=beta-gamma=frac{pi-3X}{2}$
Without lost of generality, let $BC=1$
$BD=AE=2cos(beta)=2sin(frac{X}{2})$
$BA=frac{1}{2cos(beta)}=frac{1}{2sin(frac{X}{2})}$
$CE=DA=BA-BD=frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})$
$EH=CEsin(delta)= (frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})) cos(frac{3X}{2})$
$CH=CEcos(delta)= (frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2}))sin(frac{3X}{2})$
$HD=CD-CH=1-frac{sin(frac{3X}{2} )}{2sin(frac{X}{2})}+2sin(frac{X}{2})sin(frac{3X}{2})$
$tan(Y)=tan(widehat{CDE})=frac{EH}{HD}=frac{(frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})) cos(frac{3X}{2})}{ 1-frac{sin(frac{3X}{2} )}{2sin(frac{X}{2})}+2sin(frac{X}{2})sin(frac{3X}{2}) }$
The relationship between angles $X$ and $Y$ is :
$$tan(Y)=frac{(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})}{2sin(frac{X}{2}) -sin(frac{3X}{2} )+4sin^2(frac{X}{2})sin(frac{3X}{2}) } $$
$$cot(Y)=frac{2sin(frac{X}{2}) } {(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})} -tan(frac{3X}{2}) tag 1$$
The function $Y(X)$ is represented on the next figure.
For example, if we set $X=pifrac{40}{180}$ that is an angle of $40°$ , the numerical result is :
$$Ysimeq 0.872664625997165quadtoquad 180frac{Y}{pi}simeq 50.00000000000003$$
Thus, from numerical calculus, at $Y=50°$ corresponds $X=40°$ with a very good accuracy.
Analytically one have to prove that the equation $(1)$ returns exactly $Y=pifrac{50}{180}$ for $X=pifrac{40}{180}$ .
Consider la function :
$$f(X,Y)=frac{2sin(frac{X}{2}) } {(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})} -tan(frac{3X}{2}) -cot(Y) $$
we have to prove that this function is $=0$ when $Y=pifrac{50}{180}$ and $X=pifrac{40}{180}$
$$f(frac{2pi}{9},frac{5pi}{18}) =frac{2sin(frac{pi}{9}) } {(1-4sin^2(frac{pi}{9})) cos(frac{pi}{3})} -tan(frac{pi}{3}) -cot(frac{5pi}{18}) $$
$$f((frac{2pi}{9},frac{5pi}{18}) =frac{4sin(frac{pi}{9}) } {(1-4sin^2(frac{pi}{9}))} -sqrt{3} -cot(frac{5pi}{18}) $$
Let $s=sin(frac{pi}{18})$
$sin(frac{pi}{9})=2ssqrt{1-s^2}$
$cos(frac{5pi}{18})= (16s^4-12s^2+1)sqrt{1-s^2}$
$sin(frac{5pi}{18})=(16s^5-20s^3+5s)$
$$f(frac{2pi}{9},frac{5pi}{18}) =frac{8ssqrt{1-s^2} } {1-4(2ssqrt{1-s^2})^2} -sqrt{3} -frac {(16s^4-12s^2+1)sqrt{1-s^2}}{16s^5-20s^3+5s} $$
In fact, we have to check if one of the roots of the equation :
$$frac{8ssqrt{1-s^2} } {1-4(2ssqrt{1-s^2})^2} -sqrt{3} -frac {(16s^4-12s^2+1)sqrt{1-s^2}}{16s^5-20s^3+5s}=0 tag 2$$
is equal to $sin(frac{pi}{18})$ .
Expanding and factoring transform the equation to :
$$(8s^3-6s-1)(8s^3-6s+1)(4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1)=0$$
For $s=sin(frac{pi}{18})$ the polynomials $8s^3-6s+1neq 0$ and $4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1neq 0$ .
Only the equation :
$$8s^3-6s+1=0$$
is likely to have a root around to $sin(frac{pi}{18})$. Then we have to prove that it is the case.
We put into the equation $s=sin(frac{pi}{18})=frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i}$
$$8s^3-6s+1=8left(frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i} right)^3-6left(frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i} right)+1$$
After expansion and simplification all terms cancel. We get : $8s^3-6s+1=0$ .
Thus $s=sin(frac{pi}{18})$ is a root of equation $(2)$ and as a consequence $f(frac{2pi}{9},frac{5pi}{18}) =0$ ,
This is the analytical proof that $X=40°$ and $Y=50°$ is an exact solution.
answered Feb 4 at 10:35
JJacquelinJJacquelin
45.7k21858
$endgroup$
With the notations on the figure :
$X=widehat{BAC}$
$beta=widehat{CBA}=frac{pi-X}{2}$
$gamma=widehat{DCB}=pi-2beta=X$
$delta=widehat{ACD}=beta-gamma=frac{pi-3X}{2}$
Without lost of generality, let $BC=1$
$BD=AE=2cos(beta)=2sin(frac{X}{2})$
$BA=frac{1}{2cos(beta)}=frac{1}{2sin(frac{X}{2})}$
$CE=DA=BA-BD=frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})$
$EH=CEsin(delta)= (frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})) cos(frac{3X}{2})$
$CH=CEcos(delta)= (frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2}))sin(frac{3X}{2})$
$HD=CD-CH=1-frac{sin(frac{3X}{2} )}{2sin(frac{X}{2})}+2sin(frac{X}{2})sin(frac{3X}{2})$
$tan(Y)=tan(widehat{CDE})=frac{EH}{HD}=frac{(frac{1}{2sin(frac{X}{2})}-2sin(frac{X}{2})) cos(frac{3X}{2})}{ 1-frac{sin(frac{3X}{2} )}{2sin(frac{X}{2})}+2sin(frac{X}{2})sin(frac{3X}{2}) }$
The relationship between angles $X$ and $Y$ is :
$$tan(Y)=frac{(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})}{2sin(frac{X}{2}) -sin(frac{3X}{2} )+4sin^2(frac{X}{2})sin(frac{3X}{2}) } $$
$$cot(Y)=frac{2sin(frac{X}{2}) } {(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})} -tan(frac{3X}{2}) tag 1$$
The function $Y(X)$ is represented on the next figure.
For example, if we set $X=pifrac{40}{180}$ that is an angle of $40°$ , the numerical result is :
$$Ysimeq 0.872664625997165quadtoquad 180frac{Y}{pi}simeq 50.00000000000003$$
Thus, from numerical calculus, at $Y=50°$ corresponds $X=40°$ with a very good accuracy.
Analytically one have to prove that the equation $(1)$ returns exactly $Y=pifrac{50}{180}$ for $X=pifrac{40}{180}$ .
Consider la function :
$$f(X,Y)=frac{2sin(frac{X}{2}) } {(1-4sin^2(frac{X}{2})) cos(frac{3X}{2})} -tan(frac{3X}{2}) -cot(Y) $$
we have to prove that this function is $=0$ when $Y=pifrac{50}{180}$ and $X=pifrac{40}{180}$
$$f(frac{2pi}{9},frac{5pi}{18}) =frac{2sin(frac{pi}{9}) } {(1-4sin^2(frac{pi}{9})) cos(frac{pi}{3})} -tan(frac{pi}{3}) -cot(frac{5pi}{18}) $$
$$f((frac{2pi}{9},frac{5pi}{18}) =frac{4sin(frac{pi}{9}) } {(1-4sin^2(frac{pi}{9}))} -sqrt{3} -cot(frac{5pi}{18}) $$
Let $s=sin(frac{pi}{18})$
$sin(frac{pi}{9})=2ssqrt{1-s^2}$
$cos(frac{5pi}{18})= (16s^4-12s^2+1)sqrt{1-s^2}$
$sin(frac{5pi}{18})=(16s^5-20s^3+5s)$
$$f(frac{2pi}{9},frac{5pi}{18}) =frac{8ssqrt{1-s^2} } {1-4(2ssqrt{1-s^2})^2} -sqrt{3} -frac {(16s^4-12s^2+1)sqrt{1-s^2}}{16s^5-20s^3+5s} $$
In fact, we have to check if one of the roots of the equation :
$$frac{8ssqrt{1-s^2} } {1-4(2ssqrt{1-s^2})^2} -sqrt{3} -frac {(16s^4-12s^2+1)sqrt{1-s^2}}{16s^5-20s^3+5s}=0 tag 2$$
is equal to $sin(frac{pi}{18})$ .
Expanding and factoring transform the equation to :
$$(8s^3-6s-1)(8s^3-6s+1)(4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1)=0$$
For $s=sin(frac{pi}{18})$ the polynomials $8s^3-6s+1neq 0$ and $4096s^{12}-13312s^{10}+16128s^8-8832s^6+2096s^4-176s^2+1neq 0$ .
Only the equation :
$$8s^3-6s+1=0$$
is likely to have a root around to $sin(frac{pi}{18})$. Then we have to prove that it is the case.
We put into the equation $s=sin(frac{pi}{18})=frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i}$
$$8s^3-6s+1=8left(frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i} right)^3-6left(frac{exp(ifrac{pi}{18})-exp(-ifrac{pi}{18})}{2i} right)+1$$
After expansion and simplification all terms cancel. We get : $8s^3-6s+1=0$ .
Thus $s=sin(frac{pi}{18})$ is a root of equation $(2)$ and as a consequence $f(frac{2pi}{9},frac{5pi}{18}) =0$ ,
This is the analytical proof that $X=40°$ and $Y=50°$ is an exact solution.
answered Feb 4 at 10:35
JJacquelinJJacquelin
45.7k21858
edited Feb 4 at 16:57
answered Feb 4 at 10:35
JJacquelinJJacquelin
45.7k21858
answered Feb 4 at 10:35
JJacquelinJJacquelin
45.7k21858
answered Feb 4 at 10:35
JJacquelinJJacquelin
45.7k21858
45.7k21858
add a comment |
add a comment |
$begingroup$
Here is a synthetic solution.
Let $angle BAC = x$ and $angle DCE = y$
Now, $angle ABC = 60 + frac{y}{3}$. If we prove $y= 30$, the proof would be complete. Let $BC = a$
Note that $$BD = 2a cos ({60 +frac{y}{3}})$$
and since $AB= frac {a}{2cos ({60 + frac{y}{3}})}$, we can calculate after simplifying and using the property
:$frac {4cos^2 t -1}{2cos t} = frac {3- 4sin^2 t}{2cos t}= frac {sin 3t}{sin 2t}$
$$frac {AD}{a} = frac {sin y}{cos ({30 + frac {2y}{3}})}$$
By sine rule in $Delta DEC$, and using the above result,
$$sin{y}sin({y+50})= sin{50}cos({30+frac{2y}{3}})$$
And so,
$$2sin{y}sin({y+50})= 2sin{50}cos({30+frac{2y}{3}})$$
if $y=30$, then this equation is satisfied because RHS becomes $sin 100$ and LHS becomes $sin 80$ and they are obviously equal.
Therefore, $x=40$
Note:
To see that that equation has no other solutions than $y=30$, I suggest using Wolfram Alpha.
answered Feb 5 at 11:32
DhvanitDhvanit
10010
$endgroup$
add a comment |
$begingroup$
Here is a synthetic solution.
Let $angle BAC = x$ and $angle DCE = y$
Now, $angle ABC = 60 + frac{y}{3}$. If we prove $y= 30$, the proof would be complete. Let $BC = a$
Note that $$BD = 2a cos ({60 +frac{y}{3}})$$
and since $AB= frac {a}{2cos ({60 + frac{y}{3}})}$, we can calculate after simplifying and using the property
:$frac {4cos^2 t -1}{2cos t} = frac {3- 4sin^2 t}{2cos t}= frac {sin 3t}{sin 2t}$
$$frac {AD}{a} = frac {sin y}{cos ({30 + frac {2y}{3}})}$$
By sine rule in $Delta DEC$, and using the above result,
$$sin{y}sin({y+50})= sin{50}cos({30+frac{2y}{3}})$$
And so,
$$2sin{y}sin({y+50})= 2sin{50}cos({30+frac{2y}{3}})$$
if $y=30$, then this equation is satisfied because RHS becomes $sin 100$ and LHS becomes $sin 80$ and they are obviously equal.
Therefore, $x=40$
Note:
To see that that equation has no other solutions than $y=30$, I suggest using Wolfram Alpha.
answered Feb 5 at 11:32
DhvanitDhvanit
10010
$endgroup$
add a comment |
$begingroup$
Here is a synthetic solution.
Let $angle BAC = x$ and $angle DCE = y$
Now, $angle ABC = 60 + frac{y}{3}$. If we prove $y= 30$, the proof would be complete. Let $BC = a$
Note that $$BD = 2a cos ({60 +frac{y}{3}})$$
and since $AB= frac {a}{2cos ({60 + frac{y}{3}})}$, we can calculate after simplifying and using the property
:$frac {4cos^2 t -1}{2cos t} = frac {3- 4sin^2 t}{2cos t}= frac {sin 3t}{sin 2t}$
$$frac {AD}{a} = frac {sin y}{cos ({30 + frac {2y}{3}})}$$
By sine rule in $Delta DEC$, and using the above result,
$$sin{y}sin({y+50})= sin{50}cos({30+frac{2y}{3}})$$
And so,
$$2sin{y}sin({y+50})= 2sin{50}cos({30+frac{2y}{3}})$$
if $y=30$, then this equation is satisfied because RHS becomes $sin 100$ and LHS becomes $sin 80$ and they are obviously equal.
Therefore, $x=40$
Note:
To see that that equation has no other solutions than $y=30$, I suggest using Wolfram Alpha.
answered Feb 5 at 11:32
DhvanitDhvanit
10010
$endgroup$
Here is a synthetic solution.
Let $angle BAC = x$ and $angle DCE = y$
Now, $angle ABC = 60 + frac{y}{3}$. If we prove $y= 30$, the proof would be complete. Let $BC = a$
Note that $$BD = 2a cos ({60 +frac{y}{3}})$$
and since $AB= frac {a}{2cos ({60 + frac{y}{3}})}$, we can calculate after simplifying and using the property
:$frac {4cos^2 t -1}{2cos t} = frac {3- 4sin^2 t}{2cos t}= frac {sin 3t}{sin 2t}$
$$frac {AD}{a} = frac {sin y}{cos ({30 + frac {2y}{3}})}$$
By sine rule in $Delta DEC$, and using the above result,
$$sin{y}sin({y+50})= sin{50}cos({30+frac{2y}{3}})$$
And so,
$$2sin{y}sin({y+50})= 2sin{50}cos({30+frac{2y}{3}})$$
if $y=30$, then this equation is satisfied because RHS becomes $sin 100$ and LHS becomes $sin 80$ and they are obviously equal.
Therefore, $x=40$
Note:
To see that that equation has no other solutions than $y=30$, I suggest using Wolfram Alpha.
answered Feb 5 at 11:32
DhvanitDhvanit
10010
edited Feb 5 at 11:50
answered Feb 5 at 11:32
DhvanitDhvanit
10010
answered Feb 5 at 11:32
DhvanitDhvanit
10010
answered Feb 5 at 11:32
DhvanitDhvanit
10010
10010
add a comment |
add a comment |
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I got $$x=frac{pi}{3}$$, i this right?
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– Dr. Sonnhard Graubner
Feb 3 at 13:19
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actually you will find that you can scale the diagram and that the answer will be x=40 degrees, but what my question really is " how to find it mathematically"
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– Khaled Oqab
Feb 3 at 13:42
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Ok, i will check my answer, thank you for the information!
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– Dr. Sonnhard Graubner
Feb 3 at 13:49