Mixing
How can people assume this angle is exactly half of the other angle?
$begingroup$
Long story short what I don't understand is underlined here in red:
So, they somehow seem to assume the angle on the triangle on the right has an angle $frac{theta}{2}$.
How do they know that? How can they assume it is exactly half of the angle of theta?
Thanks
EDIT: to give a bit more of a context, it has to do with the mapping of a rocket's position in the ground-fixed coordinates to the spherical earth's coordinates. When going from one coordinate to the other the altitude varies.
geometry
asked Jan 5 at 16:17
traduceradtraducerad
1184
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add a comment |
$begingroup$
Long story short what I don't understand is underlined here in red:
So, they somehow seem to assume the angle on the triangle on the right has an angle $frac{theta}{2}$.
How do they know that? How can they assume it is exactly half of the angle of theta?
Thanks
EDIT: to give a bit more of a context, it has to do with the mapping of a rocket's position in the ground-fixed coordinates to the spherical earth's coordinates. When going from one coordinate to the other the altitude varies.
geometry
asked Jan 5 at 16:17
traduceradtraducerad
1184
$endgroup$
$begingroup$
How was the triangle on the left constructed?
$endgroup$
– Mike
Jan 5 at 16:22
$begingroup$
Is this something related to physics?
$endgroup$
– Thomas Shelby
Jan 5 at 16:24
$begingroup$
@ThomasShelby yes this is related to physics
$endgroup$
– traducerad
Jan 5 at 16:32
add a comment |
$begingroup$
Long story short what I don't understand is underlined here in red:
So, they somehow seem to assume the angle on the triangle on the right has an angle $frac{theta}{2}$.
How do they know that? How can they assume it is exactly half of the angle of theta?
Thanks
EDIT: to give a bit more of a context, it has to do with the mapping of a rocket's position in the ground-fixed coordinates to the spherical earth's coordinates. When going from one coordinate to the other the altitude varies.
geometry
asked Jan 5 at 16:17
traduceradtraducerad
1184
$endgroup$
Long story short what I don't understand is underlined here in red:
So, they somehow seem to assume the angle on the triangle on the right has an angle $frac{theta}{2}$.
How do they know that? How can they assume it is exactly half of the angle of theta?
Thanks
EDIT: to give a bit more of a context, it has to do with the mapping of a rocket's position in the ground-fixed coordinates to the spherical earth's coordinates. When going from one coordinate to the other the altitude varies.
geometry
geometry
asked Jan 5 at 16:17
traduceradtraducerad
1184
asked Jan 5 at 16:17
traduceradtraducerad
1184
edited Jan 5 at 16:25
traducerad
asked Jan 5 at 16:17
traduceradtraducerad
1184
asked Jan 5 at 16:17
traduceradtraducerad
1184
asked Jan 5 at 16:17
traduceradtraducerad
1184
1184
$begingroup$
How was the triangle on the left constructed?
$endgroup$
– Mike
Jan 5 at 16:22
$begingroup$
Is this something related to physics?
$endgroup$
– Thomas Shelby
Jan 5 at 16:24
$begingroup$
@ThomasShelby yes this is related to physics
$endgroup$
– traducerad
Jan 5 at 16:32
add a comment |
$begingroup$
How was the triangle on the left constructed?
$endgroup$
– Mike
Jan 5 at 16:22
$begingroup$
Is this something related to physics?
$endgroup$
– Thomas Shelby
Jan 5 at 16:24
$begingroup$
@ThomasShelby yes this is related to physics
$endgroup$
– traducerad
Jan 5 at 16:32
$begingroup$
How was the triangle on the left constructed?
$endgroup$
– Mike
Jan 5 at 16:22
$begingroup$
How was the triangle on the left constructed?
$endgroup$
– Mike
Jan 5 at 16:22
$begingroup$
Is this something related to physics?
$endgroup$
– Thomas Shelby
Jan 5 at 16:24
$begingroup$
Is this something related to physics?
$endgroup$
– Thomas Shelby
Jan 5 at 16:24
$begingroup$
@ThomasShelby yes this is related to physics
$endgroup$
– traducerad
Jan 5 at 16:32
$begingroup$
@ThomasShelby yes this is related to physics
$endgroup$
– traducerad
Jan 5 at 16:32
add a comment |
1 Answer
1
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There is an isoscales triangle between the circle center $C$ and the 2 points $A$ and $B$ where the lines through the center meet the circle. The angle at $C$ is $theta$, so each base angle is $90^circ-fractheta2 = angle CAB$. Since the tangent has a right angle with the radius ($angle CAD=90^circ$), the angle under consideration ($angle BAD)$ is $fractheta2$.
answered Jan 5 at 16:26
IngixIngix
3,474146
$endgroup$
$begingroup$
If you could provide a small ugly drawing to clarify what you mean, I'll accept your answer
$endgroup$
– traducerad
Jan 5 at 16:27
1
$begingroup$
@traducerad Added the picture
$endgroup$
– Ingix
Jan 5 at 16:41
$begingroup$
@ThomasShelby You think the assumption that (in my notation) $CA$ is a radius and $AD$ a tangent is strange?
$endgroup$
– Ingix
Jan 5 at 16:43
$begingroup$
Great answer @Ingix
$endgroup$
– Mike
Jan 5 at 16:46
1
$begingroup$
Great question @traducerad btw!
$endgroup$
– Mike
Jan 5 at 16:54
|
show 2 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is an isoscales triangle between the circle center $C$ and the 2 points $A$ and $B$ where the lines through the center meet the circle. The angle at $C$ is $theta$, so each base angle is $90^circ-fractheta2 = angle CAB$. Since the tangent has a right angle with the radius ($angle CAD=90^circ$), the angle under consideration ($angle BAD)$ is $fractheta2$.
answered Jan 5 at 16:26
IngixIngix
3,474146
$endgroup$
$begingroup$
If you could provide a small ugly drawing to clarify what you mean, I'll accept your answer
$endgroup$
– traducerad
Jan 5 at 16:27
1
$begingroup$
@traducerad Added the picture
$endgroup$
– Ingix
Jan 5 at 16:41
$begingroup$
@ThomasShelby You think the assumption that (in my notation) $CA$ is a radius and $AD$ a tangent is strange?
$endgroup$
– Ingix
Jan 5 at 16:43
$begingroup$
Great answer @Ingix
$endgroup$
– Mike
Jan 5 at 16:46
1
$begingroup$
Great question @traducerad btw!
$endgroup$
– Mike
Jan 5 at 16:54
|
show 2 more comments
$begingroup$
There is an isoscales triangle between the circle center $C$ and the 2 points $A$ and $B$ where the lines through the center meet the circle. The angle at $C$ is $theta$, so each base angle is $90^circ-fractheta2 = angle CAB$. Since the tangent has a right angle with the radius ($angle CAD=90^circ$), the angle under consideration ($angle BAD)$ is $fractheta2$.
answered Jan 5 at 16:26
IngixIngix
3,474146
$endgroup$
$begingroup$
If you could provide a small ugly drawing to clarify what you mean, I'll accept your answer
$endgroup$
– traducerad
Jan 5 at 16:27
1
$begingroup$
@traducerad Added the picture
$endgroup$
– Ingix
Jan 5 at 16:41
$begingroup$
@ThomasShelby You think the assumption that (in my notation) $CA$ is a radius and $AD$ a tangent is strange?
$endgroup$
– Ingix
Jan 5 at 16:43
$begingroup$
Great answer @Ingix
$endgroup$
– Mike
Jan 5 at 16:46
1
$begingroup$
Great question @traducerad btw!
$endgroup$
– Mike
Jan 5 at 16:54
|
show 2 more comments
$begingroup$
There is an isoscales triangle between the circle center $C$ and the 2 points $A$ and $B$ where the lines through the center meet the circle. The angle at $C$ is $theta$, so each base angle is $90^circ-fractheta2 = angle CAB$. Since the tangent has a right angle with the radius ($angle CAD=90^circ$), the angle under consideration ($angle BAD)$ is $fractheta2$.
answered Jan 5 at 16:26
IngixIngix
3,474146
$endgroup$
There is an isoscales triangle between the circle center $C$ and the 2 points $A$ and $B$ where the lines through the center meet the circle. The angle at $C$ is $theta$, so each base angle is $90^circ-fractheta2 = angle CAB$. Since the tangent has a right angle with the radius ($angle CAD=90^circ$), the angle under consideration ($angle BAD)$ is $fractheta2$.
answered Jan 5 at 16:26
IngixIngix
3,474146
edited Jan 5 at 16:40
answered Jan 5 at 16:26
IngixIngix
3,474146
answered Jan 5 at 16:26
IngixIngix
3,474146
answered Jan 5 at 16:26
IngixIngix
3,474146
3,474146
$begingroup$
If you could provide a small ugly drawing to clarify what you mean, I'll accept your answer
$endgroup$
– traducerad
Jan 5 at 16:27
1
$begingroup$
@traducerad Added the picture
$endgroup$
– Ingix
Jan 5 at 16:41
$begingroup$
@ThomasShelby You think the assumption that (in my notation) $CA$ is a radius and $AD$ a tangent is strange?
$endgroup$
– Ingix
Jan 5 at 16:43
$begingroup$
Great answer @Ingix
$endgroup$
– Mike
Jan 5 at 16:46
1
$begingroup$
Great question @traducerad btw!
$endgroup$
– Mike
Jan 5 at 16:54
|
show 2 more comments
$begingroup$
If you could provide a small ugly drawing to clarify what you mean, I'll accept your answer
$endgroup$
– traducerad
Jan 5 at 16:27
1
$begingroup$
@traducerad Added the picture
$endgroup$
– Ingix
Jan 5 at 16:41
$begingroup$
@ThomasShelby You think the assumption that (in my notation) $CA$ is a radius and $AD$ a tangent is strange?
$endgroup$
– Ingix
Jan 5 at 16:43
$begingroup$
Great answer @Ingix
$endgroup$
– Mike
Jan 5 at 16:46
1
$begingroup$
Great question @traducerad btw!
$endgroup$
– Mike
Jan 5 at 16:54
$begingroup$
If you could provide a small ugly drawing to clarify what you mean, I'll accept your answer
$endgroup$
– traducerad
Jan 5 at 16:27
$begingroup$
If you could provide a small ugly drawing to clarify what you mean, I'll accept your answer
$endgroup$
– traducerad
Jan 5 at 16:27
1
1
$begingroup$
@traducerad Added the picture
$endgroup$
– Ingix
Jan 5 at 16:41
$begingroup$
@traducerad Added the picture
$endgroup$
– Ingix
Jan 5 at 16:41
$begingroup$
@ThomasShelby You think the assumption that (in my notation) $CA$ is a radius and $AD$ a tangent is strange?
$endgroup$
– Ingix
Jan 5 at 16:43
$begingroup$
@ThomasShelby You think the assumption that (in my notation) $CA$ is a radius and $AD$ a tangent is strange?
$endgroup$
– Ingix
Jan 5 at 16:43
$begingroup$
Great answer @Ingix
$endgroup$
– Mike
Jan 5 at 16:46
$begingroup$
Great answer @Ingix
$endgroup$
– Mike
Jan 5 at 16:46
1
1
$begingroup$
Great question @traducerad btw!
$endgroup$
– Mike
Jan 5 at 16:54
$begingroup$
Great question @traducerad btw!
$endgroup$
– Mike
Jan 5 at 16:54
|
show 2 more comments
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$begingroup$
How was the triangle on the left constructed?
$endgroup$
– Mike
Jan 5 at 16:22
$begingroup$
Is this something related to physics?
$endgroup$
– Thomas Shelby
Jan 5 at 16:24
$begingroup$
@ThomasShelby yes this is related to physics
$endgroup$
– traducerad
Jan 5 at 16:32