Using the Law of Cosines results in an “invalid” answer, why?
up vote
4
down vote
favorite
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 at 21:21
daedsidog
1374
add a comment |
up vote
4
down vote
favorite
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 at 21:21
daedsidog
1374
4
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
– Jens
Nov 26 at 21:39
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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 at 21:21
daedsidog
1374
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 at 21:21
daedsidog
1374
asked Nov 26 at 21:21
daedsidog
1374
edited Nov 26 at 21:55
asked Nov 26 at 21:21
daedsidog
1374
asked Nov 26 at 21:21
daedsidog
1374
asked Nov 26 at 21:21
daedsidog
1374
1374
4
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
– Jens
Nov 26 at 21:39
add a comment |
4
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
– Jens
Nov 26 at 21:39
4
4
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
– Jens
Nov 26 at 21:39
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
– Jens
Nov 26 at 21:39
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
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 at 21:37
Doug M
43k31753
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.
– Jason DeVito
Nov 26 at 21:42
I understand. Thank you.
– daedsidog
Nov 26 at 21:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
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 at 21:37
Doug M
43k31753
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.
– Jason DeVito
Nov 26 at 21:42
I understand. Thank you.
– daedsidog
Nov 26 at 21:43
add a comment |
up vote
5
down vote
accepted
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 at 21:37
Doug M
43k31753
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.
– Jason DeVito
Nov 26 at 21:42
I understand. Thank you.
– daedsidog
Nov 26 at 21:43
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
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 at 21:37
Doug M
43k31753
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 at 21:37
Doug M
43k31753
answered Nov 26 at 21:37
Doug M
43k31753
answered Nov 26 at 21:37
Doug M
43k31753
answered Nov 26 at 21:37
Doug M
43k31753
43k31753
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.
– Jason DeVito
Nov 26 at 21:42
I understand. Thank you.
– daedsidog
Nov 26 at 21:43
add a comment |
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.
– Jason DeVito
Nov 26 at 21:42
I understand. Thank you.
– daedsidog
Nov 26 at 21:43
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.
– Jason DeVito
Nov 26 at 21:42
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.
– Jason DeVito
Nov 26 at 21:42
I understand. Thank you.
– daedsidog
Nov 26 at 21:43
I understand. Thank you.
– daedsidog
Nov 26 at 21:43
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014945%2fusing-the-law-of-cosines-results-in-an-invalid-answer-why%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
4
The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^circ - v$.
– Jens
Nov 26 at 21:39