Mixing
Simpler Methods Geometry
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In $△ABC$, $∠A = 2∠B$ , $CD$ bisects $∠ACB$ , $AC =11cm$ and $AD = 2cm$. Find the length, in cm, of $BC$. Attached is the figure of the question
I already solved it using the Angle Bisector Theorem and sine and cosine law, but is pretty tedious to do. I am asking for solutions that would be a lot quicker.
Solution:
My system of equations:
$ 11^2 +2^2 -44cos2A = x^2 + frac4x^2121 - frac4x^2cos A11$
$BD = frac2BC11$
$fracsin2ACD = fracsin CAD11$
$fracsin ACD = fracsin CDBx$
Solving these systems would give us 13.
geometry
edited Jul 18 at 18:48
GuySa
402313
asked Jul 18 at 15:00
SuperMage1
687210
add a comment |Â
up vote
0
down vote
favorite
In $△ABC$, $∠A = 2∠B$ , $CD$ bisects $∠ACB$ , $AC =11cm$ and $AD = 2cm$. Find the length, in cm, of $BC$. Attached is the figure of the question
I already solved it using the Angle Bisector Theorem and sine and cosine law, but is pretty tedious to do. I am asking for solutions that would be a lot quicker.
Solution:
My system of equations:
$ 11^2 +2^2 -44cos2A = x^2 + frac4x^2121 - frac4x^2cos A11$
$BD = frac2BC11$
$fracsin2ACD = fracsin CAD11$
$fracsin ACD = fracsin CDBx$
Solving these systems would give us 13.
geometry
edited Jul 18 at 18:48
GuySa
402313
asked Jul 18 at 15:00
SuperMage1
687210
Could you attach the figure for this question? It's not clear what is $D$.
– Jaroslaw Matlak
Jul 18 at 15:13
Try Stewart's Theorem.
– MalayTheDynamo
Jul 18 at 15:19
Im trying but what would you do after getting CD in terms of BC?
– SuperMage1
Jul 18 at 15:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In $△ABC$, $∠A = 2∠B$ , $CD$ bisects $∠ACB$ , $AC =11cm$ and $AD = 2cm$. Find the length, in cm, of $BC$. Attached is the figure of the question
I already solved it using the Angle Bisector Theorem and sine and cosine law, but is pretty tedious to do. I am asking for solutions that would be a lot quicker.
Solution:
My system of equations:
$ 11^2 +2^2 -44cos2A = x^2 + frac4x^2121 - frac4x^2cos A11$
$BD = frac2BC11$
$fracsin2ACD = fracsin CAD11$
$fracsin ACD = fracsin CDBx$
Solving these systems would give us 13.
geometry
edited Jul 18 at 18:48
GuySa
402313
asked Jul 18 at 15:00
SuperMage1
687210
In $△ABC$, $∠A = 2∠B$ , $CD$ bisects $∠ACB$ , $AC =11cm$ and $AD = 2cm$. Find the length, in cm, of $BC$. Attached is the figure of the question
I already solved it using the Angle Bisector Theorem and sine and cosine law, but is pretty tedious to do. I am asking for solutions that would be a lot quicker.
Solution:
My system of equations:
$ 11^2 +2^2 -44cos2A = x^2 + frac4x^2121 - frac4x^2cos A11$
$BD = frac2BC11$
$fracsin2ACD = fracsin CAD11$
$fracsin ACD = fracsin CDBx$
Solving these systems would give us 13.
geometry
edited Jul 18 at 18:48
GuySa
402313
asked Jul 18 at 15:00
SuperMage1
687210
edited Jul 18 at 18:48
GuySa
402313
edited Jul 18 at 18:48
GuySa
402313
edited Jul 18 at 18:48
GuySa
402313
402313
asked Jul 18 at 15:00
SuperMage1
687210
asked Jul 18 at 15:00
SuperMage1
687210
asked Jul 18 at 15:00
SuperMage1
687210
687210
Could you attach the figure for this question? It's not clear what is $D$.
– Jaroslaw Matlak
Jul 18 at 15:13
Try Stewart's Theorem.
– MalayTheDynamo
Jul 18 at 15:19
Im trying but what would you do after getting CD in terms of BC?
– SuperMage1
Jul 18 at 15:27
add a comment |Â
Could you attach the figure for this question? It's not clear what is $D$.
– Jaroslaw Matlak
Jul 18 at 15:13
Try Stewart's Theorem.
– MalayTheDynamo
Jul 18 at 15:19
Im trying but what would you do after getting CD in terms of BC?
– SuperMage1
Jul 18 at 15:27
Could you attach the figure for this question? It's not clear what is $D$.
– Jaroslaw Matlak
Jul 18 at 15:13
Could you attach the figure for this question? It's not clear what is $D$.
– Jaroslaw Matlak
Jul 18 at 15:13
Try Stewart's Theorem.
– MalayTheDynamo
Jul 18 at 15:19
Try Stewart's Theorem.
– MalayTheDynamo
Jul 18 at 15:19
Im trying but what would you do after getting CD in terms of BC?
– SuperMage1
Jul 18 at 15:27
Im trying but what would you do after getting CD in terms of BC?
– SuperMage1
Jul 18 at 15:27
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Click for image
This is another way to do it.
Locate a point $M$ on $BC$ , such that $MC$ equals $11$. Join $AM$ , $MD$. Let the intersection of $AM$ and $CD$ be $O$
For convenience , let $$angle ACD = angle DCB = x \∠CAD = 2y\ ∠CBD = y$$ where $DC$ is $angle$ bisector.
$triangle ACM$ is isosceles. This implies the base angles are equal and are $$frac180-2x2 = frac3y2$$
(Since $180-2x = 3y$).
Also , $triangle$s $AOC$ and $MOC$ are congruent. This further implies $triangle AOD$ and $triangle MOD$ are congruent (SAS).
$$angle OAD = 2y- frac3y2 = yover 2$$ which implies $angle OMD$ also equals $y/2$ (cpct).
Also , $DM = AD = 2$ (equal base angles)
Exterior $$angle MDB = 2*yover 2 = y$$
Therefore , $MD = MB = 2$ (equal base angles)
Therefore, the length of $$BC = MC + MB = 11+2 = 13$$
answered Jul 21 at 12:10
Rahul R
264
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Click for image
This is another way to do it.
Locate a point $M$ on $BC$ , such that $MC$ equals $11$. Join $AM$ , $MD$. Let the intersection of $AM$ and $CD$ be $O$
For convenience , let $$angle ACD = angle DCB = x \∠CAD = 2y\ ∠CBD = y$$ where $DC$ is $angle$ bisector.
$triangle ACM$ is isosceles. This implies the base angles are equal and are $$frac180-2x2 = frac3y2$$
(Since $180-2x = 3y$).
Also , $triangle$s $AOC$ and $MOC$ are congruent. This further implies $triangle AOD$ and $triangle MOD$ are congruent (SAS).
$$angle OAD = 2y- frac3y2 = yover 2$$ which implies $angle OMD$ also equals $y/2$ (cpct).
Also , $DM = AD = 2$ (equal base angles)
Exterior $$angle MDB = 2*yover 2 = y$$
Therefore , $MD = MB = 2$ (equal base angles)
Therefore, the length of $$BC = MC + MB = 11+2 = 13$$
answered Jul 21 at 12:10
Rahul R
264
add a comment |Â
up vote
1
down vote
accepted
Click for image
This is another way to do it.
Locate a point $M$ on $BC$ , such that $MC$ equals $11$. Join $AM$ , $MD$. Let the intersection of $AM$ and $CD$ be $O$
For convenience , let $$angle ACD = angle DCB = x \∠CAD = 2y\ ∠CBD = y$$ where $DC$ is $angle$ bisector.
$triangle ACM$ is isosceles. This implies the base angles are equal and are $$frac180-2x2 = frac3y2$$
(Since $180-2x = 3y$).
Also , $triangle$s $AOC$ and $MOC$ are congruent. This further implies $triangle AOD$ and $triangle MOD$ are congruent (SAS).
$$angle OAD = 2y- frac3y2 = yover 2$$ which implies $angle OMD$ also equals $y/2$ (cpct).
Also , $DM = AD = 2$ (equal base angles)
Exterior $$angle MDB = 2*yover 2 = y$$
Therefore , $MD = MB = 2$ (equal base angles)
Therefore, the length of $$BC = MC + MB = 11+2 = 13$$
answered Jul 21 at 12:10
Rahul R
264
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Click for image
This is another way to do it.
Locate a point $M$ on $BC$ , such that $MC$ equals $11$. Join $AM$ , $MD$. Let the intersection of $AM$ and $CD$ be $O$
For convenience , let $$angle ACD = angle DCB = x \∠CAD = 2y\ ∠CBD = y$$ where $DC$ is $angle$ bisector.
$triangle ACM$ is isosceles. This implies the base angles are equal and are $$frac180-2x2 = frac3y2$$
(Since $180-2x = 3y$).
Also , $triangle$s $AOC$ and $MOC$ are congruent. This further implies $triangle AOD$ and $triangle MOD$ are congruent (SAS).
$$angle OAD = 2y- frac3y2 = yover 2$$ which implies $angle OMD$ also equals $y/2$ (cpct).
Also , $DM = AD = 2$ (equal base angles)
Exterior $$angle MDB = 2*yover 2 = y$$
Therefore , $MD = MB = 2$ (equal base angles)
Therefore, the length of $$BC = MC + MB = 11+2 = 13$$
answered Jul 21 at 12:10
Rahul R
264
Click for image
This is another way to do it.
Locate a point $M$ on $BC$ , such that $MC$ equals $11$. Join $AM$ , $MD$. Let the intersection of $AM$ and $CD$ be $O$
For convenience , let $$angle ACD = angle DCB = x \∠CAD = 2y\ ∠CBD = y$$ where $DC$ is $angle$ bisector.
$triangle ACM$ is isosceles. This implies the base angles are equal and are $$frac180-2x2 = frac3y2$$
(Since $180-2x = 3y$).
Also , $triangle$s $AOC$ and $MOC$ are congruent. This further implies $triangle AOD$ and $triangle MOD$ are congruent (SAS).
$$angle OAD = 2y- frac3y2 = yover 2$$ which implies $angle OMD$ also equals $y/2$ (cpct).
Also , $DM = AD = 2$ (equal base angles)
Exterior $$angle MDB = 2*yover 2 = y$$
Therefore , $MD = MB = 2$ (equal base angles)
Therefore, the length of $$BC = MC + MB = 11+2 = 13$$
answered Jul 21 at 12:10
Rahul R
264
edited Jul 22 at 6:58
answered Jul 21 at 12:10
Rahul R
264
answered Jul 21 at 12:10
Rahul R
264
answered Jul 21 at 12:10
Rahul R
264
264
add a comment |Â
add a comment |Â
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Could you attach the figure for this question? It's not clear what is $D$.
– Jaroslaw Matlak
Jul 18 at 15:13
Try Stewart's Theorem.
– MalayTheDynamo
Jul 18 at 15:19
Im trying but what would you do after getting CD in terms of BC?
– SuperMage1
Jul 18 at 15:27