Prove the law of cosines for an obtuse angle [closed]
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The question that I am pondering is that I need to derive the law of cosines for a case in which angle A is an obtuse angle.
trigonometry
asked Mar 15 '16 at 18:56
Alexis
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closed as off-topic by Micah, Jyrki Lahtonen, Jendrik Stelzner, Brahadeesh, Sil Aug 19 at 7:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Micah, Jyrki Lahtonen, Jendrik Stelzner, Brahadeesh, Sil
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The question that I am pondering is that I need to derive the law of cosines for a case in which angle A is an obtuse angle.
trigonometry
asked Mar 15 '16 at 18:56
Alexis
61
closed as off-topic by Micah, Jyrki Lahtonen, Jendrik Stelzner, Brahadeesh, Sil Aug 19 at 7:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Micah, Jyrki Lahtonen, Jendrik Stelzner, Brahadeesh, Sil
1
A good start would be to include in your question a proof that is good when angles $A$ and $B$ are both acute, and describe what happens if you make $A$ obtuse. Then at least you could say what specific problem(s) you encountered while attempting the proof.
– David K
Mar 15 '16 at 21:57
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up vote
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The question that I am pondering is that I need to derive the law of cosines for a case in which angle A is an obtuse angle.
trigonometry
asked Mar 15 '16 at 18:56
Alexis
61
The question that I am pondering is that I need to derive the law of cosines for a case in which angle A is an obtuse angle.
trigonometry
asked Mar 15 '16 at 18:56
Alexis
61
asked Mar 15 '16 at 18:56
Alexis
61
asked Mar 15 '16 at 18:56
Alexis
61
asked Mar 15 '16 at 18:56
Alexis
61
61
closed as off-topic by Micah, Jyrki Lahtonen, Jendrik Stelzner, Brahadeesh, Sil Aug 19 at 7:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Micah, Jyrki Lahtonen, Jendrik Stelzner, Brahadeesh, Sil
closed as off-topic by Micah, Jyrki Lahtonen, Jendrik Stelzner, Brahadeesh, Sil Aug 19 at 7:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Micah, Jyrki Lahtonen, Jendrik Stelzner, Brahadeesh, Sil
1
A good start would be to include in your question a proof that is good when angles $A$ and $B$ are both acute, and describe what happens if you make $A$ obtuse. Then at least you could say what specific problem(s) you encountered while attempting the proof.
– David K
Mar 15 '16 at 21:57
add a comment |Â
1
A good start would be to include in your question a proof that is good when angles $A$ and $B$ are both acute, and describe what happens if you make $A$ obtuse. Then at least you could say what specific problem(s) you encountered while attempting the proof.
– David K
Mar 15 '16 at 21:57
1
1
A good start would be to include in your question a proof that is good when angles $A$ and $B$ are both acute, and describe what happens if you make $A$ obtuse. Then at least you could say what specific problem(s) you encountered while attempting the proof.
– David K
Mar 15 '16 at 21:57
A good start would be to include in your question a proof that is good when angles $A$ and $B$ are both acute, and describe what happens if you make $A$ obtuse. Then at least you could say what specific problem(s) you encountered while attempting the proof.
– David K
Mar 15 '16 at 21:57
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2 Answers
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Take for example the obtuse triangle
Beginning with the angle in blue (the obtuse one) and counterclockwise, call the vertices $;A,B,C;$, and call $;D;$ where the straight angle is. If we call $;alpha;$ the blue angle, we then have by Pythagoras
$$AC^2=CD^2+AD^2$$
And on the triangle $;Delta CDB;$, again by Pythagoras:
$$BC^2=BD^2+CD^2implies BC^2=BD^2+AC^2-AD^2=$$
$$=AC^2+overbrace(BD-AD)^=ABoverbrace(BD+AD)^=2AD+AB=2ACcos(180-alpha)+AB=$$
$$=AC^2+AB(-2ACcosalpha+AB)=AC^2+AB^2-2ACcdot ABcosalpha$$
answered Mar 15 '16 at 19:32
DonAntonio
173k1485219
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Draw out your generic obtuse triangle. Drop the altitude to the extended base line.
Lets make B the base, and A the line to the vertex. The extension of the base line = $A cos (180-theta) = -A cos theta$
the extended base = $B - A cos theta$
the altitude = $A sin theta$
$C^2 = (B - A cos theta)^2 + (A sin theta)^2$
Multiply that out and see if you don't get the equation you are supposed to.
answered Mar 15 '16 at 19:13
Doug M
39.4k31749
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Take for example the obtuse triangle
Beginning with the angle in blue (the obtuse one) and counterclockwise, call the vertices $;A,B,C;$, and call $;D;$ where the straight angle is. If we call $;alpha;$ the blue angle, we then have by Pythagoras
$$AC^2=CD^2+AD^2$$
And on the triangle $;Delta CDB;$, again by Pythagoras:
$$BC^2=BD^2+CD^2implies BC^2=BD^2+AC^2-AD^2=$$
$$=AC^2+overbrace(BD-AD)^=ABoverbrace(BD+AD)^=2AD+AB=2ACcos(180-alpha)+AB=$$
$$=AC^2+AB(-2ACcosalpha+AB)=AC^2+AB^2-2ACcdot ABcosalpha$$
answered Mar 15 '16 at 19:32
DonAntonio
173k1485219
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up vote
1
down vote
Take for example the obtuse triangle
Beginning with the angle in blue (the obtuse one) and counterclockwise, call the vertices $;A,B,C;$, and call $;D;$ where the straight angle is. If we call $;alpha;$ the blue angle, we then have by Pythagoras
$$AC^2=CD^2+AD^2$$
And on the triangle $;Delta CDB;$, again by Pythagoras:
$$BC^2=BD^2+CD^2implies BC^2=BD^2+AC^2-AD^2=$$
$$=AC^2+overbrace(BD-AD)^=ABoverbrace(BD+AD)^=2AD+AB=2ACcos(180-alpha)+AB=$$
$$=AC^2+AB(-2ACcosalpha+AB)=AC^2+AB^2-2ACcdot ABcosalpha$$
answered Mar 15 '16 at 19:32
DonAntonio
173k1485219
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Take for example the obtuse triangle
Beginning with the angle in blue (the obtuse one) and counterclockwise, call the vertices $;A,B,C;$, and call $;D;$ where the straight angle is. If we call $;alpha;$ the blue angle, we then have by Pythagoras
$$AC^2=CD^2+AD^2$$
And on the triangle $;Delta CDB;$, again by Pythagoras:
$$BC^2=BD^2+CD^2implies BC^2=BD^2+AC^2-AD^2=$$
$$=AC^2+overbrace(BD-AD)^=ABoverbrace(BD+AD)^=2AD+AB=2ACcos(180-alpha)+AB=$$
$$=AC^2+AB(-2ACcosalpha+AB)=AC^2+AB^2-2ACcdot ABcosalpha$$
answered Mar 15 '16 at 19:32
DonAntonio
173k1485219
Take for example the obtuse triangle
Beginning with the angle in blue (the obtuse one) and counterclockwise, call the vertices $;A,B,C;$, and call $;D;$ where the straight angle is. If we call $;alpha;$ the blue angle, we then have by Pythagoras
$$AC^2=CD^2+AD^2$$
And on the triangle $;Delta CDB;$, again by Pythagoras:
$$BC^2=BD^2+CD^2implies BC^2=BD^2+AC^2-AD^2=$$
$$=AC^2+overbrace(BD-AD)^=ABoverbrace(BD+AD)^=2AD+AB=2ACcos(180-alpha)+AB=$$
$$=AC^2+AB(-2ACcosalpha+AB)=AC^2+AB^2-2ACcdot ABcosalpha$$
answered Mar 15 '16 at 19:32
DonAntonio
173k1485219
answered Mar 15 '16 at 19:32
DonAntonio
173k1485219
answered Mar 15 '16 at 19:32
DonAntonio
173k1485219
answered Mar 15 '16 at 19:32
DonAntonio
173k1485219
173k1485219
add a comment |Â
add a comment |Â
up vote
0
down vote
Draw out your generic obtuse triangle. Drop the altitude to the extended base line.
Lets make B the base, and A the line to the vertex. The extension of the base line = $A cos (180-theta) = -A cos theta$
the extended base = $B - A cos theta$
the altitude = $A sin theta$
$C^2 = (B - A cos theta)^2 + (A sin theta)^2$
Multiply that out and see if you don't get the equation you are supposed to.
answered Mar 15 '16 at 19:13
Doug M
39.4k31749
add a comment |Â
up vote
0
down vote
Draw out your generic obtuse triangle. Drop the altitude to the extended base line.
Lets make B the base, and A the line to the vertex. The extension of the base line = $A cos (180-theta) = -A cos theta$
the extended base = $B - A cos theta$
the altitude = $A sin theta$
$C^2 = (B - A cos theta)^2 + (A sin theta)^2$
Multiply that out and see if you don't get the equation you are supposed to.
answered Mar 15 '16 at 19:13
Doug M
39.4k31749
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Draw out your generic obtuse triangle. Drop the altitude to the extended base line.
Lets make B the base, and A the line to the vertex. The extension of the base line = $A cos (180-theta) = -A cos theta$
the extended base = $B - A cos theta$
the altitude = $A sin theta$
$C^2 = (B - A cos theta)^2 + (A sin theta)^2$
Multiply that out and see if you don't get the equation you are supposed to.
answered Mar 15 '16 at 19:13
Doug M
39.4k31749
Draw out your generic obtuse triangle. Drop the altitude to the extended base line.
Lets make B the base, and A the line to the vertex. The extension of the base line = $A cos (180-theta) = -A cos theta$
the extended base = $B - A cos theta$
the altitude = $A sin theta$
$C^2 = (B - A cos theta)^2 + (A sin theta)^2$
Multiply that out and see if you don't get the equation you are supposed to.
answered Mar 15 '16 at 19:13
Doug M
39.4k31749
answered Mar 15 '16 at 19:13
Doug M
39.4k31749
answered Mar 15 '16 at 19:13
Doug M
39.4k31749
answered Mar 15 '16 at 19:13
Doug M
39.4k31749
39.4k31749
add a comment |Â
add a comment |Â
1
A good start would be to include in your question a proof that is good when angles $A$ and $B$ are both acute, and describe what happens if you make $A$ obtuse. Then at least you could say what specific problem(s) you encountered while attempting the proof.
– David K
Mar 15 '16 at 21:57