Using the Law of Cosines results in an “invalid” answer, why?
$begingroup$
I have the following isosceles triangle:
I want to find the $alpha$ angle, and I know that it is obtuse.
My first instinct was to get the length of $BM$ using the Law of Cosines, which results in two answers: a negative one and a positive one; I immediately descredited the negative one because all lengths are assumed to be positive in geometry, or so have I assumed thus far...
$$BM^2 = x^2 + 0.25x^2 - x^2cos(50)$$
$$BM approx pm 0.78x$$
From here, I thought I could easily extrapolate $alpha$ by plugging it into the Law of Sines formula, but to my surprise I did not get the correct result, $alpha approx 100.53^circ$, but $alpha - 180 approx 79.47^circ$.
$$frac{BM}{sin(50)} = frac{x}{sin(alpha)}$$
$$downarrow$$
$$frac{0.78x}{sin(50)} = frac{x}{sin(alpha)}$$
$$downarrow$$
$$sin(alpha) = frac{xcdot sin(50)}{0.78x} rightarrow alpha approx 79.16^circ$$
I assume this is because I discredited what is a valid trigonometrical answer, but why is it? Until now I have been under the impression that all lengths of geometrical shapes must be positive.
I am aware there are other methods to solve this, but I am only particularly interested in why my specific one does not behave the way I want it to.
Thanks in advance.
geometry trigonometry
asked Nov 26 '18 at 21:21
daedsidogdaedsidog
29017
$endgroup$
add a comment |
$begingroup$
I have the following isosceles triangle:
I want to find the $alpha$ angle, and I know that it is obtuse.
My first instinct was to get the length of $BM$ using the Law of Cosines, which results in two answers: a negative one and a positive one; I immediately descredited the negative one because all lengths are assumed to be positive in geometry, or so have I assumed thus far...
$$BM^2 = x^2 + 0.25x^2 - x^2cos(50)$$
$$BM approx pm 0.78x$$
From here, I thought I could easily extrapolate $alpha$ by plugging it into the Law of Sines formula, but to my surprise I did not get the correct result, $alpha approx 100.53^circ$, but $alpha - 180 approx 79.47^circ$.
$$frac{BM}{sin(50)} = frac{x}{sin(alpha)}$$
$$downarrow$$
$$frac{0.78x}{sin(50)} = frac{x}{sin(alpha)}$$
$$downarrow$$
$$sin(alpha) = frac{xcdot sin(50)}{0.78x} rightarrow alpha approx 79.16^circ$$
I assume this is because I discredited what is a valid trigonometrical answer, but why is it? Until now I have been under the impression that all lengths of geometrical shapes must be positive.
I am aware there are other methods to solve this, but I am only particularly interested in why my specific one does not behave the way I want it to.
Thanks in advance.
geometry trigonometry
asked Nov 26 '18 at 21:21
daedsidogdaedsidog
29017
$endgroup$
4
$begingroup$
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
$endgroup$
– Jens
Nov 26 '18 at 21:39
add a comment |
$begingroup$
I have the following isosceles triangle:
I want to find the $alpha$ angle, and I know that it is obtuse.
My first instinct was to get the length of $BM$ using the Law of Cosines, which results in two answers: a negative one and a positive one; I immediately descredited the negative one because all lengths are assumed to be positive in geometry, or so have I assumed thus far...
$$BM^2 = x^2 + 0.25x^2 - x^2cos(50)$$
$$BM approx pm 0.78x$$
From here, I thought I could easily extrapolate $alpha$ by plugging it into the Law of Sines formula, but to my surprise I did not get the correct result, $alpha approx 100.53^circ$, but $alpha - 180 approx 79.47^circ$.
$$frac{BM}{sin(50)} = frac{x}{sin(alpha)}$$
$$downarrow$$
$$frac{0.78x}{sin(50)} = frac{x}{sin(alpha)}$$
$$downarrow$$
$$sin(alpha) = frac{xcdot sin(50)}{0.78x} rightarrow alpha approx 79.16^circ$$
I assume this is because I discredited what is a valid trigonometrical answer, but why is it? Until now I have been under the impression that all lengths of geometrical shapes must be positive.
I am aware there are other methods to solve this, but I am only particularly interested in why my specific one does not behave the way I want it to.
Thanks in advance.
geometry trigonometry
asked Nov 26 '18 at 21:21
daedsidogdaedsidog
29017
$endgroup$
I have the following isosceles triangle:
I want to find the $alpha$ angle, and I know that it is obtuse.
My first instinct was to get the length of $BM$ using the Law of Cosines, which results in two answers: a negative one and a positive one; I immediately descredited the negative one because all lengths are assumed to be positive in geometry, or so have I assumed thus far...
$$BM^2 = x^2 + 0.25x^2 - x^2cos(50)$$
$$BM approx pm 0.78x$$
From here, I thought I could easily extrapolate $alpha$ by plugging it into the Law of Sines formula, but to my surprise I did not get the correct result, $alpha approx 100.53^circ$, but $alpha - 180 approx 79.47^circ$.
$$frac{BM}{sin(50)} = frac{x}{sin(alpha)}$$
$$downarrow$$
$$frac{0.78x}{sin(50)} = frac{x}{sin(alpha)}$$
$$downarrow$$
$$sin(alpha) = frac{xcdot sin(50)}{0.78x} rightarrow alpha approx 79.16^circ$$
I assume this is because I discredited what is a valid trigonometrical answer, but why is it? Until now I have been under the impression that all lengths of geometrical shapes must be positive.
I am aware there are other methods to solve this, but I am only particularly interested in why my specific one does not behave the way I want it to.
Thanks in advance.
geometry trigonometry
geometry trigonometry
asked Nov 26 '18 at 21:21
daedsidogdaedsidog
29017
asked Nov 26 '18 at 21:21
daedsidogdaedsidog
29017
edited Nov 26 '18 at 21:55
daedsidog
asked Nov 26 '18 at 21:21
daedsidogdaedsidog
29017
asked Nov 26 '18 at 21:21
daedsidogdaedsidog
29017
asked Nov 26 '18 at 21:21
daedsidogdaedsidog
29017
29017
4
$begingroup$
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
$endgroup$
– Jens
Nov 26 '18 at 21:39
add a comment |
4
$begingroup$
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
$endgroup$
– Jens
Nov 26 '18 at 21:39
4
4
$begingroup$
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
$endgroup$
– Jens
Nov 26 '18 at 21:39
$begingroup$
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
$endgroup$
– Jens
Nov 26 '18 at 21:39
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You have:
$sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$
There are 2 values for $alpha$ between $0$ and $180^circ$ such that $sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$
one is approximately $79.4^circ$ the other is $180-79.4approx 100.6$
The $arcsin$ function on your calculator will return an answer in the interval $[-90,90]$ and you may need to go from there to find the angle you are actually looking for.
answered Nov 26 '18 at 21:37
Doug MDoug M
44.5k31854
$endgroup$
$begingroup$
Just to drive the point home to the OP: discarding the negative answer is correct. The law of cosines works just fine, but when using $arcsin$ to solve problems it only finds one possible solution, not all possible solutions.
$endgroup$
– Jason DeVito
Nov 26 '18 at 21:42
$begingroup$
I understand. Thank you.
$endgroup$
– daedsidog
Nov 26 '18 at 21:43
add a comment |
Your Answer
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$begingroup$
You have:
$sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$
There are 2 values for $alpha$ between $0$ and $180^circ$ such that $sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$
one is approximately $79.4^circ$ the other is $180-79.4approx 100.6$
The $arcsin$ function on your calculator will return an answer in the interval $[-90,90]$ and you may need to go from there to find the angle you are actually looking for.
answered Nov 26 '18 at 21:37
Doug MDoug M
44.5k31854
$endgroup$
$begingroup$
Just to drive the point home to the OP: discarding the negative answer is correct. The law of cosines works just fine, but when using $arcsin$ to solve problems it only finds one possible solution, not all possible solutions.
$endgroup$
– Jason DeVito
Nov 26 '18 at 21:42
$begingroup$
I understand. Thank you.
$endgroup$
– daedsidog
Nov 26 '18 at 21:43
add a comment |
$begingroup$
You have:
$sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$
There are 2 values for $alpha$ between $0$ and $180^circ$ such that $sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$
one is approximately $79.4^circ$ the other is $180-79.4approx 100.6$
The $arcsin$ function on your calculator will return an answer in the interval $[-90,90]$ and you may need to go from there to find the angle you are actually looking for.
answered Nov 26 '18 at 21:37
Doug MDoug M
44.5k31854
$endgroup$
$begingroup$
Just to drive the point home to the OP: discarding the negative answer is correct. The law of cosines works just fine, but when using $arcsin$ to solve problems it only finds one possible solution, not all possible solutions.
$endgroup$
– Jason DeVito
Nov 26 '18 at 21:42
$begingroup$
I understand. Thank you.
$endgroup$
– daedsidog
Nov 26 '18 at 21:43
add a comment |
$begingroup$
You have:
$sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$
There are 2 values for $alpha$ between $0$ and $180^circ$ such that $sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$
one is approximately $79.4^circ$ the other is $180-79.4approx 100.6$
The $arcsin$ function on your calculator will return an answer in the interval $[-90,90]$ and you may need to go from there to find the angle you are actually looking for.
answered Nov 26 '18 at 21:37
Doug MDoug M
44.5k31854
$endgroup$
You have:
$sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$
There are 2 values for $alpha$ between $0$ and $180^circ$ such that $sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$
one is approximately $79.4^circ$ the other is $180-79.4approx 100.6$
The $arcsin$ function on your calculator will return an answer in the interval $[-90,90]$ and you may need to go from there to find the angle you are actually looking for.
answered Nov 26 '18 at 21:37
Doug MDoug M
44.5k31854
answered Nov 26 '18 at 21:37
Doug MDoug M
44.5k31854
answered Nov 26 '18 at 21:37
Doug MDoug M
44.5k31854
answered Nov 26 '18 at 21:37
Doug MDoug M
44.5k31854
44.5k31854
$begingroup$
Just to drive the point home to the OP: discarding the negative answer is correct. The law of cosines works just fine, but when using $arcsin$ to solve problems it only finds one possible solution, not all possible solutions.
$endgroup$
– Jason DeVito
Nov 26 '18 at 21:42
$begingroup$
I understand. Thank you.
$endgroup$
– daedsidog
Nov 26 '18 at 21:43
add a comment |
$begingroup$
Just to drive the point home to the OP: discarding the negative answer is correct. The law of cosines works just fine, but when using $arcsin$ to solve problems it only finds one possible solution, not all possible solutions.
$endgroup$
– Jason DeVito
Nov 26 '18 at 21:42
$begingroup$
I understand. Thank you.
$endgroup$
– daedsidog
Nov 26 '18 at 21:43
$begingroup$
Just to drive the point home to the OP: discarding the negative answer is correct. The law of cosines works just fine, but when using $arcsin$ to solve problems it only finds one possible solution, not all possible solutions.
$endgroup$
– Jason DeVito
Nov 26 '18 at 21:42
$begingroup$
Just to drive the point home to the OP: discarding the negative answer is correct. The law of cosines works just fine, but when using $arcsin$ to solve problems it only finds one possible solution, not all possible solutions.
$endgroup$
– Jason DeVito
Nov 26 '18 at 21:42
$begingroup$
I understand. Thank you.
$endgroup$
– daedsidog
Nov 26 '18 at 21:43
$begingroup$
I understand. Thank you.
$endgroup$
– daedsidog
Nov 26 '18 at 21:43
add a comment |
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$begingroup$
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
$endgroup$
– Jens
Nov 26 '18 at 21:39