Fdtxjr

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What is the angle of ABD?

Angle between tangents and angle subtended by radii are supplementaryHow to determine this angle in terms of other variables?Bisecting an angle doesn't lead to Trisecting?Find the angle between two chords passing through points where lines are tangent to the circleUsing deductive reasoning to determine what is wrong with this diagramShow that $angle$AXC = $angle$ACBWhat is the angle of $angle BPC$ in $triangle BPC$What is the measure of $angle AOC$?$angle AOB=75°$, $angle CBD=62°$, $angle BAD=30°$ find $angle BDA$ and $angle ABD$Compute the angle in the quadrilateral

-2

$begingroup$

Beth draws $5$ points $A$, $B$, $C$, $D$, $E$ on a circle as well as the tangent to the circle at $A$, such that all angles marked with $x$ are equal.
What is the angle of $ABD$?

geometry

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edited Mar 16 at 11:55

pointguard0

1,55211021

asked Mar 16 at 11:26

mochamocha

2

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Can you divulge where the angles marked $x$ are?
  $endgroup$
  – Chrystomath
  Mar 16 at 11:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math.SE! Your question is beginning to attract down-votes and close-votes, probably because it is simply an isolated problem statement (and one that is missing key details). Please always strive to state something of what you know about a problem and/or where you got stuck. This information helps answerers tailor their responses to best serve you, without wasting anyone's time (theirs or yours) telling you things you already know or using techniques with which you are unfamiliar. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
  $endgroup$
  – Blue
  Mar 16 at 12:19
  

  
  
  
  

add a comment | 

-2

$begingroup$

Beth draws $5$ points $A$, $B$, $C$, $D$, $E$ on a circle as well as the tangent to the circle at $A$, such that all angles marked with $x$ are equal.
What is the angle of $ABD$?

geometry

share|cite|improve this question

edited Mar 16 at 11:55

pointguard0

1,55211021

asked Mar 16 at 11:26

mochamocha

2

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Can you divulge where the angles marked $x$ are?
  $endgroup$
  – Chrystomath
  Mar 16 at 11:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math.SE! Your question is beginning to attract down-votes and close-votes, probably because it is simply an isolated problem statement (and one that is missing key details). Please always strive to state something of what you know about a problem and/or where you got stuck. This information helps answerers tailor their responses to best serve you, without wasting anyone's time (theirs or yours) telling you things you already know or using techniques with which you are unfamiliar. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
  $endgroup$
  – Blue
  Mar 16 at 12:19
  

  
  
  
  

add a comment | 

$begingroup$

Beth draws $5$ points $A$, $B$, $C$, $D$, $E$ on a circle as well as the tangent to the circle at $A$, such that all angles marked with $x$ are equal.
What is the angle of $ABD$?

geometry

share|cite|improve this question

edited Mar 16 at 11:55

pointguard0

1,55211021

asked Mar 16 at 11:26

mochamocha

2

$endgroup$

share|cite|improve this question

edited Mar 16 at 11:55

pointguard0

1,55211021

asked Mar 16 at 11:26

mochamocha

2

share|cite|improve this question

edited Mar 16 at 11:55

pointguard0

1,55211021

asked Mar 16 at 11:26

mochamocha

2

edited Mar 16 at 11:55

pointguard0

1,55211021

edited Mar 16 at 11:55

pointguard0

1,55211021

asked Mar 16 at 11:26

mochamocha

2

asked Mar 16 at 11:26

mochamocha

2

2 Answers
2

active

oldest

votes

1

$begingroup$

It is hard with no diagram or knowing which angles are marked $x$, though in my imagination you could have something like this, where $x=frac180^circ5=36^circ$ and all the other shown angles are $x$ or multiples of $x$

In particular $angle ABD = 72^circ$

Added: You can easily show that a regular pentagon has all five $x$s equal using the inscribed angle theorem. Since the $x$s determine the shape of the pentagon, no irregular pentagon can have this property. This extends easily to other polygons and for $n$ sides and $n$ equal $x$s you would have $x=frac180^circn$ and $angle ABD = (n-3)x = fracn-3n 180^circ$

share|cite|improve this answer

edited Mar 16 at 16:14

answered Mar 16 at 12:27

HenryHenry

101k482169

$endgroup$

add a comment | 

0

$begingroup$

You are referring to a regular pentagon. Angle subtnded at center is $ dfrac3605=72^circ$ and angle subtented at circle boundary is half that angle $ =36^circ$

share|cite|improve this answer

answered Mar 16 at 16:16

NarasimhamNarasimham

21.1k62158

$endgroup$

add a comment | 

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1

$begingroup$

It is hard with no diagram or knowing which angles are marked $x$, though in my imagination you could have something like this, where $x=frac180^circ5=36^circ$ and all the other shown angles are $x$ or multiples of $x$

In particular $angle ABD = 72^circ$

Added: You can easily show that a regular pentagon has all five $x$s equal using the inscribed angle theorem. Since the $x$s determine the shape of the pentagon, no irregular pentagon can have this property. This extends easily to other polygons and for $n$ sides and $n$ equal $x$s you would have $x=frac180^circn$ and $angle ABD = (n-3)x = fracn-3n 180^circ$

share|cite|improve this answer

edited Mar 16 at 16:14

answered Mar 16 at 12:27

HenryHenry

101k482169

$endgroup$

add a comment | 

1

$begingroup$

It is hard with no diagram or knowing which angles are marked $x$, though in my imagination you could have something like this, where $x=frac180^circ5=36^circ$ and all the other shown angles are $x$ or multiples of $x$

In particular $angle ABD = 72^circ$

Added: You can easily show that a regular pentagon has all five $x$s equal using the inscribed angle theorem. Since the $x$s determine the shape of the pentagon, no irregular pentagon can have this property. This extends easily to other polygons and for $n$ sides and $n$ equal $x$s you would have $x=frac180^circn$ and $angle ABD = (n-3)x = fracn-3n 180^circ$

share|cite|improve this answer

edited Mar 16 at 16:14

answered Mar 16 at 12:27

HenryHenry

101k482169

$endgroup$

add a comment | 

$begingroup$

It is hard with no diagram or knowing which angles are marked $x$, though in my imagination you could have something like this, where $x=frac180^circ5=36^circ$ and all the other shown angles are $x$ or multiples of $x$

In particular $angle ABD = 72^circ$

Added: You can easily show that a regular pentagon has all five $x$s equal using the inscribed angle theorem. Since the $x$s determine the shape of the pentagon, no irregular pentagon can have this property. This extends easily to other polygons and for $n$ sides and $n$ equal $x$s you would have $x=frac180^circn$ and $angle ABD = (n-3)x = fracn-3n 180^circ$

share|cite|improve this answer

edited Mar 16 at 16:14

answered Mar 16 at 12:27

HenryHenry

101k482169

$endgroup$

share|cite|improve this answer

edited Mar 16 at 16:14

answered Mar 16 at 12:27

HenryHenry

101k482169

answered Mar 16 at 12:27

HenryHenry

101k482169

answered Mar 16 at 12:27

HenryHenry

101k482169

0

$begingroup$

You are referring to a regular pentagon. Angle subtnded at center is $ dfrac3605=72^circ$ and angle subtented at circle boundary is half that angle $ =36^circ$

share|cite|improve this answer

answered Mar 16 at 16:16

NarasimhamNarasimham

21.1k62158

$endgroup$

add a comment | 

0

$begingroup$

You are referring to a regular pentagon. Angle subtnded at center is $ dfrac3605=72^circ$ and angle subtented at circle boundary is half that angle $ =36^circ$

share|cite|improve this answer

answered Mar 16 at 16:16

NarasimhamNarasimham

21.1k62158

$endgroup$

add a comment | 

$begingroup$

You are referring to a regular pentagon. Angle subtnded at center is $ dfrac3605=72^circ$ and angle subtented at circle boundary is half that angle $ =36^circ$

share|cite|improve this answer

answered Mar 16 at 16:16

NarasimhamNarasimham

21.1k62158

$endgroup$

share|cite|improve this answer

answered Mar 16 at 16:16

NarasimhamNarasimham

21.1k62158

answered Mar 16 at 16:16

NarasimhamNarasimham

21.1k62158

answered Mar 16 at 16:16

NarasimhamNarasimham

21.1k62158