Mixing
Really don't understand how to find this angle
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How to find this angle?
Can't think of any way.
geometry
edited Nov 17 at 14:47
Robert Z
90.1k1056128
asked Nov 17 at 14:13
Abhinov Singh
723
add a comment |
up vote
0
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favorite
How to find this angle?
Can't think of any way.
geometry
edited Nov 17 at 14:47
Robert Z
90.1k1056128
asked Nov 17 at 14:13
Abhinov Singh
723
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to find this angle?
Can't think of any way.
geometry
edited Nov 17 at 14:47
Robert Z
90.1k1056128
asked Nov 17 at 14:13
Abhinov Singh
723
How to find this angle?
Can't think of any way.
geometry
geometry
edited Nov 17 at 14:47
Robert Z
90.1k1056128
asked Nov 17 at 14:13
Abhinov Singh
723
edited Nov 17 at 14:47
Robert Z
90.1k1056128
asked Nov 17 at 14:13
Abhinov Singh
723
edited Nov 17 at 14:47
Robert Z
90.1k1056128
edited Nov 17 at 14:47
Robert Z
90.1k1056128
edited Nov 17 at 14:47
Robert Z
90.1k1056128
90.1k1056128
asked Nov 17 at 14:13
Abhinov Singh
723
asked Nov 17 at 14:13
Abhinov Singh
723
asked Nov 17 at 14:13
Abhinov Singh
723
723
add a comment |
add a comment |
4 Answers
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3
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Hint. By the Law of sines applied to the $triangle DEF$,
$$frac{|DE|}{sin(angle DFE)}=frac{|DF|}{sin(angle DEF)}.$$
We have that $|DF|=R$ and $angle DEF=2alpha$. Find $|DE|$ and $angle DFE$.
Can you take it from here? Now you have "a way". Show your effort!
answered Nov 17 at 14:23
Robert Z
90.1k1056128
add a comment |
up vote
1
down vote
Some hints for the remaining calculations up to $cos(2alpha)$
- What is the nature of triangle DEA ? --> find DE
- It verifies $2cos(alpha)sin(2alpha)=cos(3alpha)$
- Simplify by $cos(alpha)$ and the resulting expression only has trigonometric lines of $2alpha$.
- This leads to solving $8x^2-4x-3=0$
Note that the figure helps discarding unwanted solutions.
answered Nov 17 at 15:22
zwim
11.2k628
add a comment |
up vote
0
down vote
Apply the law of sines to $Delta AED$ and $Delta DEF$:
$$begin{cases}frac{DE}{sin angle DAE}=2R\ frac{DF}{sin 2alpha}=frac{DE}{sin (90^circ-3alpha)}end{cases} Rightarrow begin{cases}DE=2Rcos alpha \ frac{R}{sin 2alpha}=frac{2Rcos alpha }{cos 3alpha}end{cases} Rightarrow \
cos 3alpha=2sin 2alpha cos alpha Rightarrow \
cosalpha cos 2alpha-sin alpha sin 2alpha=2sin 2alpha cos alpha Rightarrow \
cos 2alpha-2sin^2 alpha=2sin 2alpha Rightarrow \
2cos^2 alpha-1-2sin^2 alpha=2sin 2alpha Rightarrow \
2cos 2alpha-2sin 2alpha =1 Rightarrow \
frac{sqrt{2}}{2}cos 2alpha -frac{sqrt{2}}{2}sin 2alpha=frac{sqrt{2}}{4}\
cos (2alpha +45^circ)=frac{sqrt{2}}{4} Rightarrow alpha approx 12.15^circ Rightarrow \
cos 2alpha approx 0.669.$$
answered 2 days ago
farruhota
17.6k2736
add a comment |
up vote
0
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Wlog take radius of circle as unity. (in place of 2)
Take DF positive x-axis, DA as negative y-axis and next take projection of $DE$ to $DA$ and again along the $(x,y)$ axes as coordinates of a circle with origin at D.
$$ DE = DA cos alpha, x_E=DE sin alpha;, y_E=DE cos alpha;, $$
$$ (x_E,y_E) = - 2(sin alpha cos alpha ,cos^2 alpha) tag1 $$
For point $F, x_F=1, y_F=0$ and slope of EF=
$$ dfrac{2c^2}{1+2 s,c }= dfrac{1+cos 2 alpha}{1+ sin 2 alpha} $$
using Weierstrass trig half angle relations with $ t= tan alpha$
$$ dfrac{1+(1-t^2)/(1+t^2)}{1+2t/(1+t^2)}=dfrac{2}{(1+t)^2} tag2 $$
This is the same as reciprocal of $ tan 3alphatag3 $
as it makes angle $3 alpha $ external angle with $AD$ when cutting between $(D,A)$
$$ dfrac{1-3 t^2}{3t-t^3} tag4$$
Equate them, cross multiplying and simplifying you get a fourth order trigonometric polynomial having 4 solutions with two real roots.
$$ 3 t^4 +6 t^3 +2 t^2 +4 t -1 =0 tag5 $$
Results
First solution
$$alpha_1= 2.0341 = -63.82 ^{circ},quad cos 2 alpha_1= -0.6107$$
Second solution
$$ alpha_2= 0.21181 =11.5989 ^{circ}, quad cos 2alpha_2 = 0.9141 $$
The one you sketched is nearer to $ alpha approx 12^{circ}$ second case that I also verified with a construction. Hope you would sketch the second solution also.
EDIT1:
An error in coefficient of $t^3$ is noticed and corrected along with error regarding analytical solution..shown possible by farruhota.
$$ 3 t^4 + 4 t^3 +2 t^2 + 4 t -1 =0 tag5 $$
Accordingly
$$alpha_1= -57.1476 ^{circ},quad cos 2 alpha_1= -0.411438 $$
$$alpha_2= 12.1476 ^{circ}, quad cos 2alpha_2 = 0.911438. $$
answered Nov 18 at 13:02
Narasimham
20.4k52158
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Hint. By the Law of sines applied to the $triangle DEF$,
$$frac{|DE|}{sin(angle DFE)}=frac{|DF|}{sin(angle DEF)}.$$
We have that $|DF|=R$ and $angle DEF=2alpha$. Find $|DE|$ and $angle DFE$.
Can you take it from here? Now you have "a way". Show your effort!
answered Nov 17 at 14:23
Robert Z
90.1k1056128
add a comment |
up vote
3
down vote
Hint. By the Law of sines applied to the $triangle DEF$,
$$frac{|DE|}{sin(angle DFE)}=frac{|DF|}{sin(angle DEF)}.$$
We have that $|DF|=R$ and $angle DEF=2alpha$. Find $|DE|$ and $angle DFE$.
Can you take it from here? Now you have "a way". Show your effort!
answered Nov 17 at 14:23
Robert Z
90.1k1056128
add a comment |
up vote
3
down vote
up vote
3
down vote
Hint. By the Law of sines applied to the $triangle DEF$,
$$frac{|DE|}{sin(angle DFE)}=frac{|DF|}{sin(angle DEF)}.$$
We have that $|DF|=R$ and $angle DEF=2alpha$. Find $|DE|$ and $angle DFE$.
Can you take it from here? Now you have "a way". Show your effort!
answered Nov 17 at 14:23
Robert Z
90.1k1056128
Hint. By the Law of sines applied to the $triangle DEF$,
$$frac{|DE|}{sin(angle DFE)}=frac{|DF|}{sin(angle DEF)}.$$
We have that $|DF|=R$ and $angle DEF=2alpha$. Find $|DE|$ and $angle DFE$.
Can you take it from here? Now you have "a way". Show your effort!
answered Nov 17 at 14:23
Robert Z
90.1k1056128
edited Nov 17 at 14:51
answered Nov 17 at 14:23
Robert Z
90.1k1056128
answered Nov 17 at 14:23
Robert Z
90.1k1056128
answered Nov 17 at 14:23
Robert Z
90.1k1056128
90.1k1056128
add a comment |
add a comment |
up vote
1
down vote
Some hints for the remaining calculations up to $cos(2alpha)$
- What is the nature of triangle DEA ? --> find DE
- It verifies $2cos(alpha)sin(2alpha)=cos(3alpha)$
- Simplify by $cos(alpha)$ and the resulting expression only has trigonometric lines of $2alpha$.
- This leads to solving $8x^2-4x-3=0$
Note that the figure helps discarding unwanted solutions.
answered Nov 17 at 15:22
zwim
11.2k628
add a comment |
up vote
1
down vote
Some hints for the remaining calculations up to $cos(2alpha)$
- What is the nature of triangle DEA ? --> find DE
- It verifies $2cos(alpha)sin(2alpha)=cos(3alpha)$
- Simplify by $cos(alpha)$ and the resulting expression only has trigonometric lines of $2alpha$.
- This leads to solving $8x^2-4x-3=0$
Note that the figure helps discarding unwanted solutions.
answered Nov 17 at 15:22
zwim
11.2k628
add a comment |
up vote
1
down vote
up vote
1
down vote
Some hints for the remaining calculations up to $cos(2alpha)$
- What is the nature of triangle DEA ? --> find DE
- It verifies $2cos(alpha)sin(2alpha)=cos(3alpha)$
- Simplify by $cos(alpha)$ and the resulting expression only has trigonometric lines of $2alpha$.
- This leads to solving $8x^2-4x-3=0$
Note that the figure helps discarding unwanted solutions.
answered Nov 17 at 15:22
zwim
11.2k628
Some hints for the remaining calculations up to $cos(2alpha)$
- What is the nature of triangle DEA ? --> find DE
- It verifies $2cos(alpha)sin(2alpha)=cos(3alpha)$
- Simplify by $cos(alpha)$ and the resulting expression only has trigonometric lines of $2alpha$.
- This leads to solving $8x^2-4x-3=0$
Note that the figure helps discarding unwanted solutions.
answered Nov 17 at 15:22
zwim
11.2k628
answered Nov 17 at 15:22
zwim
11.2k628
answered Nov 17 at 15:22
zwim
11.2k628
answered Nov 17 at 15:22
zwim
11.2k628
11.2k628
add a comment |
add a comment |
up vote
0
down vote
Apply the law of sines to $Delta AED$ and $Delta DEF$:
$$begin{cases}frac{DE}{sin angle DAE}=2R\ frac{DF}{sin 2alpha}=frac{DE}{sin (90^circ-3alpha)}end{cases} Rightarrow begin{cases}DE=2Rcos alpha \ frac{R}{sin 2alpha}=frac{2Rcos alpha }{cos 3alpha}end{cases} Rightarrow \
cos 3alpha=2sin 2alpha cos alpha Rightarrow \
cosalpha cos 2alpha-sin alpha sin 2alpha=2sin 2alpha cos alpha Rightarrow \
cos 2alpha-2sin^2 alpha=2sin 2alpha Rightarrow \
2cos^2 alpha-1-2sin^2 alpha=2sin 2alpha Rightarrow \
2cos 2alpha-2sin 2alpha =1 Rightarrow \
frac{sqrt{2}}{2}cos 2alpha -frac{sqrt{2}}{2}sin 2alpha=frac{sqrt{2}}{4}\
cos (2alpha +45^circ)=frac{sqrt{2}}{4} Rightarrow alpha approx 12.15^circ Rightarrow \
cos 2alpha approx 0.669.$$
answered 2 days ago
farruhota
17.6k2736
add a comment |
up vote
0
down vote
Apply the law of sines to $Delta AED$ and $Delta DEF$:
$$begin{cases}frac{DE}{sin angle DAE}=2R\ frac{DF}{sin 2alpha}=frac{DE}{sin (90^circ-3alpha)}end{cases} Rightarrow begin{cases}DE=2Rcos alpha \ frac{R}{sin 2alpha}=frac{2Rcos alpha }{cos 3alpha}end{cases} Rightarrow \
cos 3alpha=2sin 2alpha cos alpha Rightarrow \
cosalpha cos 2alpha-sin alpha sin 2alpha=2sin 2alpha cos alpha Rightarrow \
cos 2alpha-2sin^2 alpha=2sin 2alpha Rightarrow \
2cos^2 alpha-1-2sin^2 alpha=2sin 2alpha Rightarrow \
2cos 2alpha-2sin 2alpha =1 Rightarrow \
frac{sqrt{2}}{2}cos 2alpha -frac{sqrt{2}}{2}sin 2alpha=frac{sqrt{2}}{4}\
cos (2alpha +45^circ)=frac{sqrt{2}}{4} Rightarrow alpha approx 12.15^circ Rightarrow \
cos 2alpha approx 0.669.$$
answered 2 days ago
farruhota
17.6k2736
add a comment |
up vote
0
down vote
up vote
0
down vote
Apply the law of sines to $Delta AED$ and $Delta DEF$:
$$begin{cases}frac{DE}{sin angle DAE}=2R\ frac{DF}{sin 2alpha}=frac{DE}{sin (90^circ-3alpha)}end{cases} Rightarrow begin{cases}DE=2Rcos alpha \ frac{R}{sin 2alpha}=frac{2Rcos alpha }{cos 3alpha}end{cases} Rightarrow \
cos 3alpha=2sin 2alpha cos alpha Rightarrow \
cosalpha cos 2alpha-sin alpha sin 2alpha=2sin 2alpha cos alpha Rightarrow \
cos 2alpha-2sin^2 alpha=2sin 2alpha Rightarrow \
2cos^2 alpha-1-2sin^2 alpha=2sin 2alpha Rightarrow \
2cos 2alpha-2sin 2alpha =1 Rightarrow \
frac{sqrt{2}}{2}cos 2alpha -frac{sqrt{2}}{2}sin 2alpha=frac{sqrt{2}}{4}\
cos (2alpha +45^circ)=frac{sqrt{2}}{4} Rightarrow alpha approx 12.15^circ Rightarrow \
cos 2alpha approx 0.669.$$
answered 2 days ago
farruhota
17.6k2736
Apply the law of sines to $Delta AED$ and $Delta DEF$:
$$begin{cases}frac{DE}{sin angle DAE}=2R\ frac{DF}{sin 2alpha}=frac{DE}{sin (90^circ-3alpha)}end{cases} Rightarrow begin{cases}DE=2Rcos alpha \ frac{R}{sin 2alpha}=frac{2Rcos alpha }{cos 3alpha}end{cases} Rightarrow \
cos 3alpha=2sin 2alpha cos alpha Rightarrow \
cosalpha cos 2alpha-sin alpha sin 2alpha=2sin 2alpha cos alpha Rightarrow \
cos 2alpha-2sin^2 alpha=2sin 2alpha Rightarrow \
2cos^2 alpha-1-2sin^2 alpha=2sin 2alpha Rightarrow \
2cos 2alpha-2sin 2alpha =1 Rightarrow \
frac{sqrt{2}}{2}cos 2alpha -frac{sqrt{2}}{2}sin 2alpha=frac{sqrt{2}}{4}\
cos (2alpha +45^circ)=frac{sqrt{2}}{4} Rightarrow alpha approx 12.15^circ Rightarrow \
cos 2alpha approx 0.669.$$
answered 2 days ago
farruhota
17.6k2736
answered 2 days ago
farruhota
17.6k2736
answered 2 days ago
farruhota
17.6k2736
answered 2 days ago
farruhota
17.6k2736
17.6k2736
add a comment |
add a comment |
up vote
0
down vote
Wlog take radius of circle as unity. (in place of 2)
Take DF positive x-axis, DA as negative y-axis and next take projection of $DE$ to $DA$ and again along the $(x,y)$ axes as coordinates of a circle with origin at D.
$$ DE = DA cos alpha, x_E=DE sin alpha;, y_E=DE cos alpha;, $$
$$ (x_E,y_E) = - 2(sin alpha cos alpha ,cos^2 alpha) tag1 $$
For point $F, x_F=1, y_F=0$ and slope of EF=
$$ dfrac{2c^2}{1+2 s,c }= dfrac{1+cos 2 alpha}{1+ sin 2 alpha} $$
using Weierstrass trig half angle relations with $ t= tan alpha$
$$ dfrac{1+(1-t^2)/(1+t^2)}{1+2t/(1+t^2)}=dfrac{2}{(1+t)^2} tag2 $$
This is the same as reciprocal of $ tan 3alphatag3 $
as it makes angle $3 alpha $ external angle with $AD$ when cutting between $(D,A)$
$$ dfrac{1-3 t^2}{3t-t^3} tag4$$
Equate them, cross multiplying and simplifying you get a fourth order trigonometric polynomial having 4 solutions with two real roots.
$$ 3 t^4 +6 t^3 +2 t^2 +4 t -1 =0 tag5 $$
Results
First solution
$$alpha_1= 2.0341 = -63.82 ^{circ},quad cos 2 alpha_1= -0.6107$$
Second solution
$$ alpha_2= 0.21181 =11.5989 ^{circ}, quad cos 2alpha_2 = 0.9141 $$
The one you sketched is nearer to $ alpha approx 12^{circ}$ second case that I also verified with a construction. Hope you would sketch the second solution also.
EDIT1:
An error in coefficient of $t^3$ is noticed and corrected along with error regarding analytical solution..shown possible by farruhota.
$$ 3 t^4 + 4 t^3 +2 t^2 + 4 t -1 =0 tag5 $$
Accordingly
$$alpha_1= -57.1476 ^{circ},quad cos 2 alpha_1= -0.411438 $$
$$alpha_2= 12.1476 ^{circ}, quad cos 2alpha_2 = 0.911438. $$
answered Nov 18 at 13:02
Narasimham
20.4k52158
add a comment |
up vote
0
down vote
Wlog take radius of circle as unity. (in place of 2)
Take DF positive x-axis, DA as negative y-axis and next take projection of $DE$ to $DA$ and again along the $(x,y)$ axes as coordinates of a circle with origin at D.
$$ DE = DA cos alpha, x_E=DE sin alpha;, y_E=DE cos alpha;, $$
$$ (x_E,y_E) = - 2(sin alpha cos alpha ,cos^2 alpha) tag1 $$
For point $F, x_F=1, y_F=0$ and slope of EF=
$$ dfrac{2c^2}{1+2 s,c }= dfrac{1+cos 2 alpha}{1+ sin 2 alpha} $$
using Weierstrass trig half angle relations with $ t= tan alpha$
$$ dfrac{1+(1-t^2)/(1+t^2)}{1+2t/(1+t^2)}=dfrac{2}{(1+t)^2} tag2 $$
This is the same as reciprocal of $ tan 3alphatag3 $
as it makes angle $3 alpha $ external angle with $AD$ when cutting between $(D,A)$
$$ dfrac{1-3 t^2}{3t-t^3} tag4$$
Equate them, cross multiplying and simplifying you get a fourth order trigonometric polynomial having 4 solutions with two real roots.
$$ 3 t^4 +6 t^3 +2 t^2 +4 t -1 =0 tag5 $$
Results
First solution
$$alpha_1= 2.0341 = -63.82 ^{circ},quad cos 2 alpha_1= -0.6107$$
Second solution
$$ alpha_2= 0.21181 =11.5989 ^{circ}, quad cos 2alpha_2 = 0.9141 $$
The one you sketched is nearer to $ alpha approx 12^{circ}$ second case that I also verified with a construction. Hope you would sketch the second solution also.
EDIT1:
An error in coefficient of $t^3$ is noticed and corrected along with error regarding analytical solution..shown possible by farruhota.
$$ 3 t^4 + 4 t^3 +2 t^2 + 4 t -1 =0 tag5 $$
Accordingly
$$alpha_1= -57.1476 ^{circ},quad cos 2 alpha_1= -0.411438 $$
$$alpha_2= 12.1476 ^{circ}, quad cos 2alpha_2 = 0.911438. $$
answered Nov 18 at 13:02
Narasimham
20.4k52158
add a comment |
up vote
0
down vote
up vote
0
down vote
Wlog take radius of circle as unity. (in place of 2)
Take DF positive x-axis, DA as negative y-axis and next take projection of $DE$ to $DA$ and again along the $(x,y)$ axes as coordinates of a circle with origin at D.
$$ DE = DA cos alpha, x_E=DE sin alpha;, y_E=DE cos alpha;, $$
$$ (x_E,y_E) = - 2(sin alpha cos alpha ,cos^2 alpha) tag1 $$
For point $F, x_F=1, y_F=0$ and slope of EF=
$$ dfrac{2c^2}{1+2 s,c }= dfrac{1+cos 2 alpha}{1+ sin 2 alpha} $$
using Weierstrass trig half angle relations with $ t= tan alpha$
$$ dfrac{1+(1-t^2)/(1+t^2)}{1+2t/(1+t^2)}=dfrac{2}{(1+t)^2} tag2 $$
This is the same as reciprocal of $ tan 3alphatag3 $
as it makes angle $3 alpha $ external angle with $AD$ when cutting between $(D,A)$
$$ dfrac{1-3 t^2}{3t-t^3} tag4$$
Equate them, cross multiplying and simplifying you get a fourth order trigonometric polynomial having 4 solutions with two real roots.
$$ 3 t^4 +6 t^3 +2 t^2 +4 t -1 =0 tag5 $$
Results
First solution
$$alpha_1= 2.0341 = -63.82 ^{circ},quad cos 2 alpha_1= -0.6107$$
Second solution
$$ alpha_2= 0.21181 =11.5989 ^{circ}, quad cos 2alpha_2 = 0.9141 $$
The one you sketched is nearer to $ alpha approx 12^{circ}$ second case that I also verified with a construction. Hope you would sketch the second solution also.
EDIT1:
An error in coefficient of $t^3$ is noticed and corrected along with error regarding analytical solution..shown possible by farruhota.
$$ 3 t^4 + 4 t^3 +2 t^2 + 4 t -1 =0 tag5 $$
Accordingly
$$alpha_1= -57.1476 ^{circ},quad cos 2 alpha_1= -0.411438 $$
$$alpha_2= 12.1476 ^{circ}, quad cos 2alpha_2 = 0.911438. $$
answered Nov 18 at 13:02
Narasimham
20.4k52158
Wlog take radius of circle as unity. (in place of 2)
Take DF positive x-axis, DA as negative y-axis and next take projection of $DE$ to $DA$ and again along the $(x,y)$ axes as coordinates of a circle with origin at D.
$$ DE = DA cos alpha, x_E=DE sin alpha;, y_E=DE cos alpha;, $$
$$ (x_E,y_E) = - 2(sin alpha cos alpha ,cos^2 alpha) tag1 $$
For point $F, x_F=1, y_F=0$ and slope of EF=
$$ dfrac{2c^2}{1+2 s,c }= dfrac{1+cos 2 alpha}{1+ sin 2 alpha} $$
using Weierstrass trig half angle relations with $ t= tan alpha$
$$ dfrac{1+(1-t^2)/(1+t^2)}{1+2t/(1+t^2)}=dfrac{2}{(1+t)^2} tag2 $$
This is the same as reciprocal of $ tan 3alphatag3 $
as it makes angle $3 alpha $ external angle with $AD$ when cutting between $(D,A)$
$$ dfrac{1-3 t^2}{3t-t^3} tag4$$
Equate them, cross multiplying and simplifying you get a fourth order trigonometric polynomial having 4 solutions with two real roots.
$$ 3 t^4 +6 t^3 +2 t^2 +4 t -1 =0 tag5 $$
Results
First solution
$$alpha_1= 2.0341 = -63.82 ^{circ},quad cos 2 alpha_1= -0.6107$$
Second solution
$$ alpha_2= 0.21181 =11.5989 ^{circ}, quad cos 2alpha_2 = 0.9141 $$
The one you sketched is nearer to $ alpha approx 12^{circ}$ second case that I also verified with a construction. Hope you would sketch the second solution also.
EDIT1:
An error in coefficient of $t^3$ is noticed and corrected along with error regarding analytical solution..shown possible by farruhota.
$$ 3 t^4 + 4 t^3 +2 t^2 + 4 t -1 =0 tag5 $$
Accordingly
$$alpha_1= -57.1476 ^{circ},quad cos 2 alpha_1= -0.411438 $$
$$alpha_2= 12.1476 ^{circ}, quad cos 2alpha_2 = 0.911438. $$
answered Nov 18 at 13:02
Narasimham
20.4k52158
edited yesterday
answered Nov 18 at 13:02
Narasimham
20.4k52158
answered Nov 18 at 13:02
Narasimham
20.4k52158
answered Nov 18 at 13:02
Narasimham
20.4k52158
20.4k52158
add a comment |
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