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- In case of right angled triangles, C^2 = A^2 +B^2 - 2AB cos(Ψ) is shortened to C^2 = A^2 +B^2 because the cosine of the angle "Ψ" which is 90° is equal to 0. But how is its cosine equal to 0.
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Not Correct.But how is its cosine equal to 0.
You can find the proof of Law Of Cosines in some Trigonometry books and some Calculus books. There is an old book by Anton which includes a proof.Summary:: In case of right angled triangles, C^2 = A^2 +B^2 - 2AB cos(Ψ) is shortened to C^2 = A^2 +B^2 because the cosine of the angle "Ψ" which is 90° is equal to 0. But how is its cosine equal to 0.
IView attachment 264987
Summary:: In case of right angled triangles, C^2 = A^2 +B^2 - 2AB cos(Ψ) is shortened to C^2 = A^2 +B^2 because the cosine of the angle "Ψ" which is 90° is equal to 0. But how is its cosine equal to 0.
cos 90 = sin(90-90) = sin(0) = 0.
Does that now beg the question "how is sin(0) = 0" ?
😁 Oh you English people! I can see that you have not even used the scale for drawing that perpendicular ##h## and that ##h## is very pointy, it has sharp end points. It’s really an English’s writing (so tough to read).Another way, label the triangle like this:
View attachment 265361
Pythagoras on the left triangle: ##x^2 + h^2 = b^2##
Pythagoras on the right triangle: ##a^2 - 2ax + (x^2 + h^2) = c^2##
Now since ##b\cos{C} = x##, we finally have ##a^2 + b^2 - 2ab\cos{C} = c^2##.
😁 Oh you English people! I can see that you have not even used the scale for drawing that perpendicular ##h## and that ##h## is very pointy, it has sharp end points. It’s really an English’s writing (so tough to read).
Draw an unit circle and add an additional line on the positive y-axis as the terminal arm. Let z denote the length of the terminal arm, we get: $$\cos(x)=\frac x z\implies \cos(90)=\frac 0 1 = 0$$
I'm pretty sure that he means the ray whose angle is measured relative to the reference direction, the positive x-axis.What do you mean by a terminal arm?
##h=b\sin\hat{C}## and ##x=b\cos\hat{C}##. Hence:Another way, label the triangle like this:
Pythagoras on the left triangle: ##x^2 + h^2 = b^2##
Pythagoras on the right triangle: ##a^2 - 2ax + (x^2 + h^2) = c^2##
Now since ##b\cos{C} = x##, we finally have ##a^2 + b^2 - 2ab\cos{C} = c^2##.
It is a great loss of archaeology that you do not work in that field.##h=b\sin\hat{C}## and ##x=b\cos\hat{C}##. Hence:
$$c^2=b^2\sin^2\hat{C}+(a-b\cos\hat{C})^2=a^2 + b^2 - 2ab\cos{\hat{C}}$$, and as a 'bonus': $$\tan\hat{B}=\frac{b\sin\hat{C}}{a-b\cos\hat{C}}$$It
thanks - cool profile pic by the way. Where is it from ?It is a great loss of archaeology that you do not work in that field.
I found it on a website for sharing wallpapers.