Finding the length of a side in an irregular quadrilateral, given three angle measures and two other sidesConstruct a Triangle from Given Base, Obtuse Angle Adjacent to Base and Difference of Two Other SidesFinding side and angle of isosceles triangle inside two circlesRelation between the GM of two sides of a triangle and the bisector of angle between themInside angle of a triangle with two sides of known ratio and one known side?Finding side of a triangle, given two sides and angle bisectorTwo sides of a triangle and the angle in between them are known. How to find the 3rd side?Help me find a side of a quadrilateral given one side and distance between the centers of two circles inscribed in the same quadrilateralFinding angle of irregular quadrilateral given all sides and one angleFinding an Angle using Cyclic Quadrilateral and Circle TheoremsFinding the length of the side of a triangle, given the lengths of the other two sides.
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Finding the length of a side in an irregular quadrilateral, given three angle measures and two other sides
Construct a Triangle from Given Base, Obtuse Angle Adjacent to Base and Difference of Two Other SidesFinding side and angle of isosceles triangle inside two circlesRelation between the GM of two sides of a triangle and the bisector of angle between themInside angle of a triangle with two sides of known ratio and one known side?Finding side of a triangle, given two sides and angle bisectorTwo sides of a triangle and the angle in between them are known. How to find the 3rd side?Help me find a side of a quadrilateral given one side and distance between the centers of two circles inscribed in the same quadrilateralFinding angle of irregular quadrilateral given all sides and one angleFinding an Angle using Cyclic Quadrilateral and Circle TheoremsFinding the length of the side of a triangle, given the lengths of the other two sides.
$begingroup$
I'm struggling with finding the length $x$. It is obvious that $angle DCB = 150^circ$.
geometry
edited Mar 29 at 20:49
Blue
49.5k870157
asked Mar 29 at 20:42
BobtrollstenBobtrollsten
7115
$endgroup$
add a comment |
$begingroup$
I'm struggling with finding the length $x$. It is obvious that $angle DCB = 150^circ$.
geometry
edited Mar 29 at 20:49
Blue
49.5k870157
asked Mar 29 at 20:42
BobtrollstenBobtrollsten
7115
$endgroup$
$begingroup$
Extend $AD$ and $BC$ to meet at $E$. $ABE$ is then an equilateral triangle.
$endgroup$
– FredH
Mar 29 at 20:46
add a comment |
$begingroup$
I'm struggling with finding the length $x$. It is obvious that $angle DCB = 150^circ$.
geometry
edited Mar 29 at 20:49
Blue
49.5k870157
asked Mar 29 at 20:42
BobtrollstenBobtrollsten
7115
$endgroup$
I'm struggling with finding the length $x$. It is obvious that $angle DCB = 150^circ$.
geometry
geometry
edited Mar 29 at 20:49
Blue
49.5k870157
asked Mar 29 at 20:42
BobtrollstenBobtrollsten
7115
edited Mar 29 at 20:49
Blue
49.5k870157
asked Mar 29 at 20:42
BobtrollstenBobtrollsten
7115
edited Mar 29 at 20:49
Blue
49.5k870157
edited Mar 29 at 20:49
Blue
49.5k870157
edited Mar 29 at 20:49
Blue
49.5k870157
49.5k870157
asked Mar 29 at 20:42
BobtrollstenBobtrollsten
7115
asked Mar 29 at 20:42
BobtrollstenBobtrollsten
7115
asked Mar 29 at 20:42
BobtrollstenBobtrollsten
7115
7115
$begingroup$
Extend $AD$ and $BC$ to meet at $E$. $ABE$ is then an equilateral triangle.
$endgroup$
– FredH
Mar 29 at 20:46
add a comment |
$begingroup$
Extend $AD$ and $BC$ to meet at $E$. $ABE$ is then an equilateral triangle.
$endgroup$
– FredH
Mar 29 at 20:46
$begingroup$
Extend $AD$ and $BC$ to meet at $E$. $ABE$ is then an equilateral triangle.
$endgroup$
– FredH
Mar 29 at 20:46
$begingroup$
Extend $AD$ and $BC$ to meet at $E$. $ABE$ is then an equilateral triangle.
$endgroup$
– FredH
Mar 29 at 20:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Extend $overlineAD$ and $overlineBC$ to form an equilateral triangle $triangle ABE$. Then notice that $triangle EDB$ is a $30$-$60$-$90$ right triangle whose $sqrt3$ side is $5sqrt3$. Therefore, the length of $overlineDE$ is $5$, the length of $overlineBE$ is $10$ and $x=9$.
answered Mar 29 at 20:50
John DoumaJohn Douma
5,83521520
$endgroup$
$begingroup$
Is it possible that you show it on a diagram?
$endgroup$
– Bobtrollsten
Mar 29 at 20:51
$begingroup$
@Bobtrollsten If you follow the directions given, then you will see for yourself how this works.
$endgroup$
– John Douma
Mar 29 at 20:59
add a comment |
$begingroup$
Consider $Pin AD$ such that $CPparallel AB$. Then you have that $angle DCP=30°$. Finally $$tan(angle DCP)=tan(30°)=frac1sqrt3=fracx-45sqrt3$$ Thus
$$5=x-4iff colorredx=9$$
answered Mar 29 at 20:47
Dr. MathvaDr. Mathva
3,428630
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Extend $overlineAD$ and $overlineBC$ to form an equilateral triangle $triangle ABE$. Then notice that $triangle EDB$ is a $30$-$60$-$90$ right triangle whose $sqrt3$ side is $5sqrt3$. Therefore, the length of $overlineDE$ is $5$, the length of $overlineBE$ is $10$ and $x=9$.
answered Mar 29 at 20:50
John DoumaJohn Douma
5,83521520
$endgroup$
$begingroup$
Is it possible that you show it on a diagram?
$endgroup$
– Bobtrollsten
Mar 29 at 20:51
$begingroup$
@Bobtrollsten If you follow the directions given, then you will see for yourself how this works.
$endgroup$
– John Douma
Mar 29 at 20:59
add a comment |
$begingroup$
Extend $overlineAD$ and $overlineBC$ to form an equilateral triangle $triangle ABE$. Then notice that $triangle EDB$ is a $30$-$60$-$90$ right triangle whose $sqrt3$ side is $5sqrt3$. Therefore, the length of $overlineDE$ is $5$, the length of $overlineBE$ is $10$ and $x=9$.
answered Mar 29 at 20:50
John DoumaJohn Douma
5,83521520
$endgroup$
$begingroup$
Is it possible that you show it on a diagram?
$endgroup$
– Bobtrollsten
Mar 29 at 20:51
$begingroup$
@Bobtrollsten If you follow the directions given, then you will see for yourself how this works.
$endgroup$
– John Douma
Mar 29 at 20:59
add a comment |
$begingroup$
Extend $overlineAD$ and $overlineBC$ to form an equilateral triangle $triangle ABE$. Then notice that $triangle EDB$ is a $30$-$60$-$90$ right triangle whose $sqrt3$ side is $5sqrt3$. Therefore, the length of $overlineDE$ is $5$, the length of $overlineBE$ is $10$ and $x=9$.
answered Mar 29 at 20:50
John DoumaJohn Douma
5,83521520
$endgroup$
Extend $overlineAD$ and $overlineBC$ to form an equilateral triangle $triangle ABE$. Then notice that $triangle EDB$ is a $30$-$60$-$90$ right triangle whose $sqrt3$ side is $5sqrt3$. Therefore, the length of $overlineDE$ is $5$, the length of $overlineBE$ is $10$ and $x=9$.
answered Mar 29 at 20:50
John DoumaJohn Douma
5,83521520
answered Mar 29 at 20:50
John DoumaJohn Douma
5,83521520
answered Mar 29 at 20:50
John DoumaJohn Douma
5,83521520
answered Mar 29 at 20:50
John DoumaJohn Douma
5,83521520
5,83521520
$begingroup$
Is it possible that you show it on a diagram?
$endgroup$
– Bobtrollsten
Mar 29 at 20:51
$begingroup$
@Bobtrollsten If you follow the directions given, then you will see for yourself how this works.
$endgroup$
– John Douma
Mar 29 at 20:59
add a comment |
$begingroup$
Is it possible that you show it on a diagram?
$endgroup$
– Bobtrollsten
Mar 29 at 20:51
$begingroup$
@Bobtrollsten If you follow the directions given, then you will see for yourself how this works.
$endgroup$
– John Douma
Mar 29 at 20:59
$begingroup$
Is it possible that you show it on a diagram?
$endgroup$
– Bobtrollsten
Mar 29 at 20:51
$begingroup$
Is it possible that you show it on a diagram?
$endgroup$
– Bobtrollsten
Mar 29 at 20:51
$begingroup$
@Bobtrollsten If you follow the directions given, then you will see for yourself how this works.
$endgroup$
– John Douma
Mar 29 at 20:59
$begingroup$
@Bobtrollsten If you follow the directions given, then you will see for yourself how this works.
$endgroup$
– John Douma
Mar 29 at 20:59
add a comment |
$begingroup$
Consider $Pin AD$ such that $CPparallel AB$. Then you have that $angle DCP=30°$. Finally $$tan(angle DCP)=tan(30°)=frac1sqrt3=fracx-45sqrt3$$ Thus
$$5=x-4iff colorredx=9$$
answered Mar 29 at 20:47
Dr. MathvaDr. Mathva
3,428630
$endgroup$
add a comment |
$begingroup$
Consider $Pin AD$ such that $CPparallel AB$. Then you have that $angle DCP=30°$. Finally $$tan(angle DCP)=tan(30°)=frac1sqrt3=fracx-45sqrt3$$ Thus
$$5=x-4iff colorredx=9$$
answered Mar 29 at 20:47
Dr. MathvaDr. Mathva
3,428630
$endgroup$
add a comment |
$begingroup$
Consider $Pin AD$ such that $CPparallel AB$. Then you have that $angle DCP=30°$. Finally $$tan(angle DCP)=tan(30°)=frac1sqrt3=fracx-45sqrt3$$ Thus
$$5=x-4iff colorredx=9$$
answered Mar 29 at 20:47
Dr. MathvaDr. Mathva
3,428630
$endgroup$
Consider $Pin AD$ such that $CPparallel AB$. Then you have that $angle DCP=30°$. Finally $$tan(angle DCP)=tan(30°)=frac1sqrt3=fracx-45sqrt3$$ Thus
$$5=x-4iff colorredx=9$$
answered Mar 29 at 20:47
Dr. MathvaDr. Mathva
3,428630
edited Mar 29 at 20:58
answered Mar 29 at 20:47
Dr. MathvaDr. Mathva
3,428630
answered Mar 29 at 20:47
Dr. MathvaDr. Mathva
3,428630
answered Mar 29 at 20:47
Dr. MathvaDr. Mathva
3,428630
3,428630
add a comment |
add a comment |
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$begingroup$
Extend $AD$ and $BC$ to meet at $E$. $ABE$ is then an equilateral triangle.
$endgroup$
– FredH
Mar 29 at 20:46