Calculating hypotenuses of acute triangles in a circular segment
$begingroup$
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:
- Radius of the circular segment is known.
- Angle of the sector (and hence the segments) is known.
- Lengths $UE1$ and $JE1$ are known.
$EJ$ is parallel to the X-axis.- Assume α is the angle for each segment.
My approach:
- Calculate $KJ$, $KJ = tan(α) cdot EJ$
- From here, line $EK = KJ / sin(α)$
- To find the "opposite" of the next segment I do $EJcdot(tan(2α) - tan(α))$
- 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
asked 5 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:
- Radius of the circular segment is known.
- Angle of the sector (and hence the segments) is known.
- Lengths $UE1$ and $JE1$ are known.
$EJ$ is parallel to the X-axis.- Assume α is the angle for each segment.
My approach:
- Calculate $KJ$, $KJ = tan(α) cdot EJ$
- From here, line $EK = KJ / sin(α)$
- To find the "opposite" of the next segment I do $EJcdot(tan(2α) - tan(α))$
- 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
asked 5 hours ago
ShibaliciousShibalicious
1245
$endgroup$
$begingroup$
According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
$endgroup$
– J. W. Tanner
18 mins ago
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:
- Radius of the circular segment is known.
- Angle of the sector (and hence the segments) is known.
- Lengths $UE1$ and $JE1$ are known.
$EJ$ is parallel to the X-axis.- Assume α is the angle for each segment.
My approach:
- Calculate $KJ$, $KJ = tan(α) cdot EJ$
- From here, line $EK = KJ / sin(α)$
- To find the "opposite" of the next segment I do $EJcdot(tan(2α) - tan(α))$
- 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
asked 5 hours ago
ShibaliciousShibalicious
1245
$endgroup$
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:
- Radius of the circular segment is known.
- Angle of the sector (and hence the segments) is known.
- Lengths $UE1$ and $JE1$ are known.
$EJ$ is parallel to the X-axis.- Assume α is the angle for each segment.
My approach:
- Calculate $KJ$, $KJ = tan(α) cdot EJ$
- From here, line $EK = KJ / sin(α)$
- To find the "opposite" of the next segment I do $EJcdot(tan(2α) - tan(α))$
- 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
geometry trigonometry triangle circle programming
asked 5 hours ago
ShibaliciousShibalicious
1245
asked 5 hours ago
ShibaliciousShibalicious
1245
asked 5 hours ago
ShibaliciousShibalicious
1245
asked 5 hours ago
ShibaliciousShibalicious
1245
asked 5 hours ago
ShibaliciousShibalicious
1245
1245
$begingroup$
According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
$endgroup$
– J. W. Tanner
18 mins ago
add a comment |
$begingroup$
According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
$endgroup$
– J. W. Tanner
18 mins ago
$begingroup$
According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
$endgroup$
– J. W. Tanner
18 mins ago
$begingroup$
According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
$endgroup$
– J. W. Tanner
18 mins ago
add a comment |
1 Answer
1
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votes
$begingroup$
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
$UE_1=x_0-y_0cot 16alpha$.
answered 3 hours ago
AretinoAretino
24k21443
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
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
$UE_1=x_0-y_0cot 16alpha$.
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
$UE_1=x_0-y_0cot 16alpha$.
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
$UE_1=x_0-y_0cot 16alpha$.
answered 3 hours ago
AretinoAretino
24k21443
$endgroup$
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
$UE_1=x_0-y_0cot 16alpha$.
answered 3 hours ago
AretinoAretino
24k21443
edited 3 hours ago
answered 3 hours ago
AretinoAretino
24k21443
answered 3 hours ago
AretinoAretino
24k21443
answered 3 hours ago
AretinoAretino
24k21443
24k21443
add a comment |
add a comment |
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$begingroup$
According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
$endgroup$
– J. W. Tanner
18 mins ago