Mixing
Determine angles of a triangle given lengths of its sides
$begingroup$
If I remember correctly this is high school material; I feel ashamed that I can't solve this now.
Lengths of a triangle's sides determine its angles; but how to compute these angles?
geometry triangle angle
asked Jan 11 at 15:56
gaazkamgaazkam
456314
$endgroup$
add a comment |
$begingroup$
If I remember correctly this is high school material; I feel ashamed that I can't solve this now.
Lengths of a triangle's sides determine its angles; but how to compute these angles?
geometry triangle angle
asked Jan 11 at 15:56
gaazkamgaazkam
456314
$endgroup$
$begingroup$
I would recommend the Law of Cosines, since you don't know any of the angles. Solve as follows: $dfrac{a^2+b^2-c^2}{2ab}=cos(gamma),$ where $gamma$ is the angle opposite $c$.
$endgroup$
– Adrian Keister
Jan 11 at 16:02
add a comment |
$begingroup$
If I remember correctly this is high school material; I feel ashamed that I can't solve this now.
Lengths of a triangle's sides determine its angles; but how to compute these angles?
geometry triangle angle
asked Jan 11 at 15:56
gaazkamgaazkam
456314
$endgroup$
If I remember correctly this is high school material; I feel ashamed that I can't solve this now.
Lengths of a triangle's sides determine its angles; but how to compute these angles?
geometry triangle angle
geometry triangle angle
asked Jan 11 at 15:56
gaazkamgaazkam
456314
asked Jan 11 at 15:56
gaazkamgaazkam
456314
asked Jan 11 at 15:56
gaazkamgaazkam
456314
asked Jan 11 at 15:56
gaazkamgaazkam
456314
asked Jan 11 at 15:56
gaazkamgaazkam
456314
456314
$begingroup$
I would recommend the Law of Cosines, since you don't know any of the angles. Solve as follows: $dfrac{a^2+b^2-c^2}{2ab}=cos(gamma),$ where $gamma$ is the angle opposite $c$.
$endgroup$
– Adrian Keister
Jan 11 at 16:02
add a comment |
$begingroup$
I would recommend the Law of Cosines, since you don't know any of the angles. Solve as follows: $dfrac{a^2+b^2-c^2}{2ab}=cos(gamma),$ where $gamma$ is the angle opposite $c$.
$endgroup$
– Adrian Keister
Jan 11 at 16:02
$begingroup$
I would recommend the Law of Cosines, since you don't know any of the angles. Solve as follows: $dfrac{a^2+b^2-c^2}{2ab}=cos(gamma),$ where $gamma$ is the angle opposite $c$.
$endgroup$
– Adrian Keister
Jan 11 at 16:02
$begingroup$
I would recommend the Law of Cosines, since you don't know any of the angles. Solve as follows: $dfrac{a^2+b^2-c^2}{2ab}=cos(gamma),$ where $gamma$ is the angle opposite $c$.
$endgroup$
– Adrian Keister
Jan 11 at 16:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, @ajotatxe, it's not the law of sine, it's rather the law of cosine
$$c^2=a^2+b^2-2 a b cos(gamma)$$
(which clearly can be solved for $cos(gamma)$, and thus for $gamma$.
--- rk
answered Jan 11 at 16:03
Dr. Richard KlitzingDr. Richard Klitzing
1,64516
$endgroup$
add a comment |
$begingroup$
I'm not a massive fan of the use of $gamma$ in the question, comment & answer.
To me, with the following image, although it is poorly drawn, is much easier to remember the cosine rule. It is a high-school formula after all!
We have that:
$$A^2=B^2+C^2-2BCcos(a)to a=arccosbigg(frac{B^2+C^2-A^2}{2BC}bigg)$$
$$B^2=A^2+C^2-2ACcos(b)to b=arccosbigg(frac{A^2+C^2-B^2}{2AC}bigg)$$
$$C^2=A^2+B^2-2ABcos(c)to c=arccosbigg(frac{A^2+B^2-C^2}{2AB}bigg)$$
answered Jan 11 at 16:19
Rhys HughesRhys Hughes
6,0681529
$endgroup$
1
$begingroup$
In some countries are vertices denoted by capital letters, segments or lines by small letters of Latin alphabet, and angles by Greek alphabet. Moreover, side opposite to the vertex A is a, and angle at A is $alpha.$ It is really practical. If you don't like, you've right to take other notation, but it is not a good idea to use the letters from OP in a different sense in the solution/answer.
$endgroup$
– user376343
Jan 12 at 11:54
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, @ajotatxe, it's not the law of sine, it's rather the law of cosine
$$c^2=a^2+b^2-2 a b cos(gamma)$$
(which clearly can be solved for $cos(gamma)$, and thus for $gamma$.
--- rk
answered Jan 11 at 16:03
Dr. Richard KlitzingDr. Richard Klitzing
1,64516
$endgroup$
add a comment |
$begingroup$
No, @ajotatxe, it's not the law of sine, it's rather the law of cosine
$$c^2=a^2+b^2-2 a b cos(gamma)$$
(which clearly can be solved for $cos(gamma)$, and thus for $gamma$.
--- rk
answered Jan 11 at 16:03
Dr. Richard KlitzingDr. Richard Klitzing
1,64516
$endgroup$
add a comment |
$begingroup$
No, @ajotatxe, it's not the law of sine, it's rather the law of cosine
$$c^2=a^2+b^2-2 a b cos(gamma)$$
(which clearly can be solved for $cos(gamma)$, and thus for $gamma$.
--- rk
answered Jan 11 at 16:03
Dr. Richard KlitzingDr. Richard Klitzing
1,64516
$endgroup$
No, @ajotatxe, it's not the law of sine, it's rather the law of cosine
$$c^2=a^2+b^2-2 a b cos(gamma)$$
(which clearly can be solved for $cos(gamma)$, and thus for $gamma$.
--- rk
answered Jan 11 at 16:03
Dr. Richard KlitzingDr. Richard Klitzing
1,64516
answered Jan 11 at 16:03
Dr. Richard KlitzingDr. Richard Klitzing
1,64516
answered Jan 11 at 16:03
Dr. Richard KlitzingDr. Richard Klitzing
1,64516
answered Jan 11 at 16:03
Dr. Richard KlitzingDr. Richard Klitzing
1,64516
1,64516
add a comment |
add a comment |
$begingroup$
I'm not a massive fan of the use of $gamma$ in the question, comment & answer.
To me, with the following image, although it is poorly drawn, is much easier to remember the cosine rule. It is a high-school formula after all!
We have that:
$$A^2=B^2+C^2-2BCcos(a)to a=arccosbigg(frac{B^2+C^2-A^2}{2BC}bigg)$$
$$B^2=A^2+C^2-2ACcos(b)to b=arccosbigg(frac{A^2+C^2-B^2}{2AC}bigg)$$
$$C^2=A^2+B^2-2ABcos(c)to c=arccosbigg(frac{A^2+B^2-C^2}{2AB}bigg)$$
answered Jan 11 at 16:19
Rhys HughesRhys Hughes
6,0681529
$endgroup$
1
$begingroup$
In some countries are vertices denoted by capital letters, segments or lines by small letters of Latin alphabet, and angles by Greek alphabet. Moreover, side opposite to the vertex A is a, and angle at A is $alpha.$ It is really practical. If you don't like, you've right to take other notation, but it is not a good idea to use the letters from OP in a different sense in the solution/answer.
$endgroup$
– user376343
Jan 12 at 11:54
add a comment |
$begingroup$
I'm not a massive fan of the use of $gamma$ in the question, comment & answer.
To me, with the following image, although it is poorly drawn, is much easier to remember the cosine rule. It is a high-school formula after all!
We have that:
$$A^2=B^2+C^2-2BCcos(a)to a=arccosbigg(frac{B^2+C^2-A^2}{2BC}bigg)$$
$$B^2=A^2+C^2-2ACcos(b)to b=arccosbigg(frac{A^2+C^2-B^2}{2AC}bigg)$$
$$C^2=A^2+B^2-2ABcos(c)to c=arccosbigg(frac{A^2+B^2-C^2}{2AB}bigg)$$
answered Jan 11 at 16:19
Rhys HughesRhys Hughes
6,0681529
$endgroup$
1
$begingroup$
In some countries are vertices denoted by capital letters, segments or lines by small letters of Latin alphabet, and angles by Greek alphabet. Moreover, side opposite to the vertex A is a, and angle at A is $alpha.$ It is really practical. If you don't like, you've right to take other notation, but it is not a good idea to use the letters from OP in a different sense in the solution/answer.
$endgroup$
– user376343
Jan 12 at 11:54
add a comment |
$begingroup$
I'm not a massive fan of the use of $gamma$ in the question, comment & answer.
To me, with the following image, although it is poorly drawn, is much easier to remember the cosine rule. It is a high-school formula after all!
We have that:
$$A^2=B^2+C^2-2BCcos(a)to a=arccosbigg(frac{B^2+C^2-A^2}{2BC}bigg)$$
$$B^2=A^2+C^2-2ACcos(b)to b=arccosbigg(frac{A^2+C^2-B^2}{2AC}bigg)$$
$$C^2=A^2+B^2-2ABcos(c)to c=arccosbigg(frac{A^2+B^2-C^2}{2AB}bigg)$$
answered Jan 11 at 16:19
Rhys HughesRhys Hughes
6,0681529
$endgroup$
I'm not a massive fan of the use of $gamma$ in the question, comment & answer.
To me, with the following image, although it is poorly drawn, is much easier to remember the cosine rule. It is a high-school formula after all!
We have that:
$$A^2=B^2+C^2-2BCcos(a)to a=arccosbigg(frac{B^2+C^2-A^2}{2BC}bigg)$$
$$B^2=A^2+C^2-2ACcos(b)to b=arccosbigg(frac{A^2+C^2-B^2}{2AC}bigg)$$
$$C^2=A^2+B^2-2ABcos(c)to c=arccosbigg(frac{A^2+B^2-C^2}{2AB}bigg)$$
answered Jan 11 at 16:19
Rhys HughesRhys Hughes
6,0681529
answered Jan 11 at 16:19
Rhys HughesRhys Hughes
6,0681529
answered Jan 11 at 16:19
Rhys HughesRhys Hughes
6,0681529
answered Jan 11 at 16:19
Rhys HughesRhys Hughes
6,0681529
6,0681529
1
$begingroup$
In some countries are vertices denoted by capital letters, segments or lines by small letters of Latin alphabet, and angles by Greek alphabet. Moreover, side opposite to the vertex A is a, and angle at A is $alpha.$ It is really practical. If you don't like, you've right to take other notation, but it is not a good idea to use the letters from OP in a different sense in the solution/answer.
$endgroup$
– user376343
Jan 12 at 11:54
add a comment |
1
$begingroup$
In some countries are vertices denoted by capital letters, segments or lines by small letters of Latin alphabet, and angles by Greek alphabet. Moreover, side opposite to the vertex A is a, and angle at A is $alpha.$ It is really practical. If you don't like, you've right to take other notation, but it is not a good idea to use the letters from OP in a different sense in the solution/answer.
$endgroup$
– user376343
Jan 12 at 11:54
1
1
$begingroup$
In some countries are vertices denoted by capital letters, segments or lines by small letters of Latin alphabet, and angles by Greek alphabet. Moreover, side opposite to the vertex A is a, and angle at A is $alpha.$ It is really practical. If you don't like, you've right to take other notation, but it is not a good idea to use the letters from OP in a different sense in the solution/answer.
$endgroup$
– user376343
Jan 12 at 11:54
$begingroup$
In some countries are vertices denoted by capital letters, segments or lines by small letters of Latin alphabet, and angles by Greek alphabet. Moreover, side opposite to the vertex A is a, and angle at A is $alpha.$ It is really practical. If you don't like, you've right to take other notation, but it is not a good idea to use the letters from OP in a different sense in the solution/answer.
$endgroup$
– user376343
Jan 12 at 11:54
add a comment |
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$begingroup$
I would recommend the Law of Cosines, since you don't know any of the angles. Solve as follows: $dfrac{a^2+b^2-c^2}{2ab}=cos(gamma),$ where $gamma$ is the angle opposite $c$.
$endgroup$
– Adrian Keister
Jan 11 at 16:02