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Calculating hypotenuses of acute triangles in a circular segment

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1

$begingroup$

I have a sector of a circle split into 16 equal segments. I am trying to calculate the hypotenuses of the triangles formed by 2 intersecting straight lines (for example, triangle $LEQ$), which sort encloses the circular sector in rectangular boundaries.

Conditions:

1. Radius of the circular segment is known.


2. Angle of the sector (and hence the segments) is known.


3. Lengths $UE1$ and $JE1$ are known.


4. 
   $EJ$ is parallel to the X-axis.


5. Assume α is the angle for each segment.

My approach:

1. Calculate $KJ$, $KJ = tan(α) cdot EJ$
   


2. From here, line $EK = KJ / sin(α)$
   


3. To find the "opposite" of the next segment I do $EJcdot(tan(2α) - tan(α))$
   


4. Repeat step number 2 with the new value.

The red segments' "opposites" are parallel to the Y-axis. That all works fine until I reach segment 9, where the point $E1$ is. The "opposites" for the blue segments now have to be parallel to the X-axis, and continuing with my approach I only get them parallel with the Y-axis.

Using first calculation as a reference, how can I find the blue "opposites" such as $UT, TS, OE1, NO$, etc..?

geometry trigonometry triangle circle programming

share|cite|improve this question

asked 4 hours ago

ShibaliciousShibalicious

1245

$endgroup$

add a comment | 

1

$begingroup$

I have a sector of a circle split into 16 equal segments. I am trying to calculate the hypotenuses of the triangles formed by 2 intersecting straight lines (for example, triangle $LEQ$), which sort encloses the circular sector in rectangular boundaries.

Conditions:

1. Radius of the circular segment is known.


2. Angle of the sector (and hence the segments) is known.


3. Lengths $UE1$ and $JE1$ are known.


4. 
   $EJ$ is parallel to the X-axis.


5. Assume α is the angle for each segment.

My approach:

1. Calculate $KJ$, $KJ = tan(α) cdot EJ$
   


2. From here, line $EK = KJ / sin(α)$
   


3. To find the "opposite" of the next segment I do $EJcdot(tan(2α) - tan(α))$
   


4. Repeat step number 2 with the new value.

The red segments' "opposites" are parallel to the Y-axis. That all works fine until I reach segment 9, where the point $E1$ is. The "opposites" for the blue segments now have to be parallel to the X-axis, and continuing with my approach I only get them parallel with the Y-axis.

Using first calculation as a reference, how can I find the blue "opposites" such as $UT, TS, OE1, NO$, etc..?

geometry trigonometry triangle circle programming

share|cite|improve this question

asked 4 hours ago

ShibaliciousShibalicious

1245

$endgroup$

add a comment | 

$begingroup$

I have a sector of a circle split into 16 equal segments. I am trying to calculate the hypotenuses of the triangles formed by 2 intersecting straight lines (for example, triangle $LEQ$), which sort encloses the circular sector in rectangular boundaries.

Conditions:

1. Radius of the circular segment is known.


2. Angle of the sector (and hence the segments) is known.


3. Lengths $UE1$ and $JE1$ are known.


4. 
   $EJ$ is parallel to the X-axis.


5. Assume α is the angle for each segment.

My approach:

1. Calculate $KJ$, $KJ = tan(α) cdot EJ$
   


2. From here, line $EK = KJ / sin(α)$
   


3. To find the "opposite" of the next segment I do $EJcdot(tan(2α) - tan(α))$
   


4. Repeat step number 2 with the new value.

The red segments' "opposites" are parallel to the Y-axis. That all works fine until I reach segment 9, where the point $E1$ is. The "opposites" for the blue segments now have to be parallel to the X-axis, and continuing with my approach I only get them parallel with the Y-axis.

Using first calculation as a reference, how can I find the blue "opposites" such as $UT, TS, OE1, NO$, etc..?

geometry trigonometry triangle circle programming

share|cite|improve this question

asked 4 hours ago

ShibaliciousShibalicious

1245

$endgroup$

share|cite|improve this question

asked 4 hours ago

ShibaliciousShibalicious

1245

share|cite|improve this question

asked 4 hours ago

ShibaliciousShibalicious

1245

asked 4 hours ago

ShibaliciousShibalicious

1245

asked 4 hours ago

ShibaliciousShibalicious

1245

1 Answer
1

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votes

4

$begingroup$

You then have 16 lines $r_k$, forming angles $kalpha$ with $x$ axis ($1le kle16$).

And then you have two lines $x=x_0$, $y=y_0$, parallel to the axes.

The intersections of lines $r_k$ with these can be readily found as:
$$
P_k=(x_0, x_0tan kalpha),quad Q_k=(y_0cot kalpha, y_0).
$$
In your diagram, $K=P_1$, $R=P_2$, and so on. $A_1$ corresponds to the last $P_k$ whose ordinate is less then $y_0$, that is to:
$$
bar k=leftlfloor{1overalpha}arctan{y_0over x_0}rightrfloor.
$$
In the same way, $U=Q_{16}$, $T=Q_{15}$ and so on, down to $bar k+1$.

To compute $x_0$ and $y_0$ from your data, notice that $y_0=JE_1$ and
$UE_1=x_0-y_0cot 16alpha$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

AretinoAretino

24k21443

$endgroup$

add a comment | 

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4

$begingroup$

You then have 16 lines $r_k$, forming angles $kalpha$ with $x$ axis ($1le kle16$).

And then you have two lines $x=x_0$, $y=y_0$, parallel to the axes.

The intersections of lines $r_k$ with these can be readily found as:
$$
P_k=(x_0, x_0tan kalpha),quad Q_k=(y_0cot kalpha, y_0).
$$
In your diagram, $K=P_1$, $R=P_2$, and so on. $A_1$ corresponds to the last $P_k$ whose ordinate is less then $y_0$, that is to:
$$
bar k=leftlfloor{1overalpha}arctan{y_0over x_0}rightrfloor.
$$
In the same way, $U=Q_{16}$, $T=Q_{15}$ and so on, down to $bar k+1$.

To compute $x_0$ and $y_0$ from your data, notice that $y_0=JE_1$ and
$UE_1=x_0-y_0cot 16alpha$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

AretinoAretino

24k21443

$endgroup$

add a comment | 

4

$begingroup$

You then have 16 lines $r_k$, forming angles $kalpha$ with $x$ axis ($1le kle16$).

And then you have two lines $x=x_0$, $y=y_0$, parallel to the axes.

The intersections of lines $r_k$ with these can be readily found as:
$$
P_k=(x_0, x_0tan kalpha),quad Q_k=(y_0cot kalpha, y_0).
$$
In your diagram, $K=P_1$, $R=P_2$, and so on. $A_1$ corresponds to the last $P_k$ whose ordinate is less then $y_0$, that is to:
$$
bar k=leftlfloor{1overalpha}arctan{y_0over x_0}rightrfloor.
$$
In the same way, $U=Q_{16}$, $T=Q_{15}$ and so on, down to $bar k+1$.

To compute $x_0$ and $y_0$ from your data, notice that $y_0=JE_1$ and
$UE_1=x_0-y_0cot 16alpha$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

AretinoAretino

24k21443

$endgroup$

add a comment | 

$begingroup$

You then have 16 lines $r_k$, forming angles $kalpha$ with $x$ axis ($1le kle16$).

And then you have two lines $x=x_0$, $y=y_0$, parallel to the axes.

The intersections of lines $r_k$ with these can be readily found as:
$$
P_k=(x_0, x_0tan kalpha),quad Q_k=(y_0cot kalpha, y_0).
$$
In your diagram, $K=P_1$, $R=P_2$, and so on. $A_1$ corresponds to the last $P_k$ whose ordinate is less then $y_0$, that is to:
$$
bar k=leftlfloor{1overalpha}arctan{y_0over x_0}rightrfloor.
$$
In the same way, $U=Q_{16}$, $T=Q_{15}$ and so on, down to $bar k+1$.

To compute $x_0$ and $y_0$ from your data, notice that $y_0=JE_1$ and
$UE_1=x_0-y_0cot 16alpha$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

AretinoAretino

24k21443

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

AretinoAretino

24k21443

answered 3 hours ago

AretinoAretino

24k21443

answered 3 hours ago

AretinoAretino

24k21443