Construct a Triangle from Given Base, Obtuse Angle Adjacent to Base and Difference of Two Other Sides ...
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Construct a Triangle from Given Base, Obtuse Angle Adjacent to Base and Difference of Two Other Sides
The Next CEO of Stack OverflowCongruent TrianglesCharacteristics emerging from subdividing an obtuse scalene triangle?if $AD=BD,angle ADC=3angle CAB,AB=sqrt{2},BC=sqrt{17},CD=sqrt{10}$,How find $AC$Relation between the GM of two sides of a triangle and the bisector of angle between themSolving right triangle given the area and one angleLocus of vertex given base and ratio of other two sidesconstruct triangle with $hat C$ and length of the bisector of $hat C$ and side cConstruct an equilateral triangle with area equal to a given triangleAngles and relations of in triangles and quadilateralsright angle equal to obtuse triangle?
$begingroup$
I need to construct a triangle from given base, obtuse angle adjacent to base and difference of two other sides.
Let us try to analyze the scenario.
We are given base BC, obtuse $angletext{ACB}$ adjacent to base BC,
and BD equal to AB minus AC.
We need to construct $triangletext{ABC}$.
Now in $triangletext{ADC}$, $text{AD} = text{AC}$. Hence,
$angletext{ACD} = angletext{ADC}$.
From the given information, I can draw base BC and $angletext{ACB}$.
If
I can find the value of $angletext{ACD}$, I can draw it and draw an
arc with center at B and radius equal to BD, and then construct the triangle.
However, I find no way to to calculate $angletext{ACD}$.
I understand that,
$angletext{ACD} = angletext{ADC} = angletext{DBC} + angletext{DCB}$.
But that is where my thought process stops.
Or may be I completely in the wrong direction.
Any suggestion will be appreciated.
geometry triangles
asked Mar 1 '13 at 2:20
MasroorMasroor
80231228
$endgroup$
add a comment |
$begingroup$
I need to construct a triangle from given base, obtuse angle adjacent to base and difference of two other sides.
Let us try to analyze the scenario.
We are given base BC, obtuse $angletext{ACB}$ adjacent to base BC,
and BD equal to AB minus AC.
We need to construct $triangletext{ABC}$.
Now in $triangletext{ADC}$, $text{AD} = text{AC}$. Hence,
$angletext{ACD} = angletext{ADC}$.
From the given information, I can draw base BC and $angletext{ACB}$.
If
I can find the value of $angletext{ACD}$, I can draw it and draw an
arc with center at B and radius equal to BD, and then construct the triangle.
However, I find no way to to calculate $angletext{ACD}$.
I understand that,
$angletext{ACD} = angletext{ADC} = angletext{DBC} + angletext{DCB}$.
But that is where my thought process stops.
Or may be I completely in the wrong direction.
Any suggestion will be appreciated.
geometry triangles
asked Mar 1 '13 at 2:20
MasroorMasroor
80231228
$endgroup$
add a comment |
$begingroup$
I need to construct a triangle from given base, obtuse angle adjacent to base and difference of two other sides.
Let us try to analyze the scenario.
We are given base BC, obtuse $angletext{ACB}$ adjacent to base BC,
and BD equal to AB minus AC.
We need to construct $triangletext{ABC}$.
Now in $triangletext{ADC}$, $text{AD} = text{AC}$. Hence,
$angletext{ACD} = angletext{ADC}$.
From the given information, I can draw base BC and $angletext{ACB}$.
If
I can find the value of $angletext{ACD}$, I can draw it and draw an
arc with center at B and radius equal to BD, and then construct the triangle.
However, I find no way to to calculate $angletext{ACD}$.
I understand that,
$angletext{ACD} = angletext{ADC} = angletext{DBC} + angletext{DCB}$.
But that is where my thought process stops.
Or may be I completely in the wrong direction.
Any suggestion will be appreciated.
geometry triangles
asked Mar 1 '13 at 2:20
MasroorMasroor
80231228
$endgroup$
I need to construct a triangle from given base, obtuse angle adjacent to base and difference of two other sides.
Let us try to analyze the scenario.
We are given base BC, obtuse $angletext{ACB}$ adjacent to base BC,
and BD equal to AB minus AC.
We need to construct $triangletext{ABC}$.
Now in $triangletext{ADC}$, $text{AD} = text{AC}$. Hence,
$angletext{ACD} = angletext{ADC}$.
From the given information, I can draw base BC and $angletext{ACB}$.
If
I can find the value of $angletext{ACD}$, I can draw it and draw an
arc with center at B and radius equal to BD, and then construct the triangle.
However, I find no way to to calculate $angletext{ACD}$.
I understand that,
$angletext{ACD} = angletext{ADC} = angletext{DBC} + angletext{DCB}$.
But that is where my thought process stops.
Or may be I completely in the wrong direction.
Any suggestion will be appreciated.
geometry triangles
geometry triangles
asked Mar 1 '13 at 2:20
MasroorMasroor
80231228
asked Mar 1 '13 at 2:20
MasroorMasroor
80231228
edited Mar 1 '13 at 2:33
Masroor
asked Mar 1 '13 at 2:20
MasroorMasroor
80231228
asked Mar 1 '13 at 2:20
MasroorMasroor
80231228
asked Mar 1 '13 at 2:20
MasroorMasroor
80231228
80231228
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Continue $AC$ to $E$ so that $CE=DB$. Connect $E$ to $B$. Draw $BA$ so that $ABE$ is isosceles, i.e. make base angles $AEB$ and $ABE$ to be equal.
answered Mar 1 '13 at 3:55
MaesumiMaesumi
2,6881523
$endgroup$
$begingroup$
Thanks, looks like I was thinking in the wrong direction.
$endgroup$
– Masroor
Mar 1 '13 at 4:51
$begingroup$
AEB will be an obtuse angle.Then there can't be a way to construct an isosceles triangle with another angle equal to AEB.
$endgroup$
– rah4927
Apr 16 '14 at 12:56
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Continue $AC$ to $E$ so that $CE=DB$. Connect $E$ to $B$. Draw $BA$ so that $ABE$ is isosceles, i.e. make base angles $AEB$ and $ABE$ to be equal.
answered Mar 1 '13 at 3:55
MaesumiMaesumi
2,6881523
$endgroup$
$begingroup$
Thanks, looks like I was thinking in the wrong direction.
$endgroup$
– Masroor
Mar 1 '13 at 4:51
$begingroup$
AEB will be an obtuse angle.Then there can't be a way to construct an isosceles triangle with another angle equal to AEB.
$endgroup$
– rah4927
Apr 16 '14 at 12:56
add a comment |
$begingroup$
Continue $AC$ to $E$ so that $CE=DB$. Connect $E$ to $B$. Draw $BA$ so that $ABE$ is isosceles, i.e. make base angles $AEB$ and $ABE$ to be equal.
answered Mar 1 '13 at 3:55
MaesumiMaesumi
2,6881523
$endgroup$
$begingroup$
Thanks, looks like I was thinking in the wrong direction.
$endgroup$
– Masroor
Mar 1 '13 at 4:51
$begingroup$
AEB will be an obtuse angle.Then there can't be a way to construct an isosceles triangle with another angle equal to AEB.
$endgroup$
– rah4927
Apr 16 '14 at 12:56
add a comment |
$begingroup$
Continue $AC$ to $E$ so that $CE=DB$. Connect $E$ to $B$. Draw $BA$ so that $ABE$ is isosceles, i.e. make base angles $AEB$ and $ABE$ to be equal.
answered Mar 1 '13 at 3:55
MaesumiMaesumi
2,6881523
$endgroup$
Continue $AC$ to $E$ so that $CE=DB$. Connect $E$ to $B$. Draw $BA$ so that $ABE$ is isosceles, i.e. make base angles $AEB$ and $ABE$ to be equal.
answered Mar 1 '13 at 3:55
MaesumiMaesumi
2,6881523
answered Mar 1 '13 at 3:55
MaesumiMaesumi
2,6881523
answered Mar 1 '13 at 3:55
MaesumiMaesumi
2,6881523
answered Mar 1 '13 at 3:55
MaesumiMaesumi
2,6881523
2,6881523
$begingroup$
Thanks, looks like I was thinking in the wrong direction.
$endgroup$
– Masroor
Mar 1 '13 at 4:51
$begingroup$
AEB will be an obtuse angle.Then there can't be a way to construct an isosceles triangle with another angle equal to AEB.
$endgroup$
– rah4927
Apr 16 '14 at 12:56
add a comment |
$begingroup$
Thanks, looks like I was thinking in the wrong direction.
$endgroup$
– Masroor
Mar 1 '13 at 4:51
$begingroup$
AEB will be an obtuse angle.Then there can't be a way to construct an isosceles triangle with another angle equal to AEB.
$endgroup$
– rah4927
Apr 16 '14 at 12:56
$begingroup$
Thanks, looks like I was thinking in the wrong direction.
$endgroup$
– Masroor
Mar 1 '13 at 4:51
$begingroup$
Thanks, looks like I was thinking in the wrong direction.
$endgroup$
– Masroor
Mar 1 '13 at 4:51
$begingroup$
AEB will be an obtuse angle.Then there can't be a way to construct an isosceles triangle with another angle equal to AEB.
$endgroup$
– rah4927
Apr 16 '14 at 12:56
$begingroup$
AEB will be an obtuse angle.Then there can't be a way to construct an isosceles triangle with another angle equal to AEB.
$endgroup$
– rah4927
Apr 16 '14 at 12:56
add a comment |
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