I just peek at your link, A, B, C are sides, X, Y, Z angles.

C^2 = A^2 + B^2 − 2 A B cos(Z)

So, for right angle Z, A^2 + B^2 = C^2, as expected

Edit: typo fixed

(08-03-2018 05:59 PM)Albert Chan Wrote: [ -> ]I just peek at your link, A, B, C are sides, X, Y, Z angles.

A^2 + B^2 = C^2 - 2 A B cos(Z)

I'd say A² + B² = C² + 2AB cos(Z)

Or C² = A² + B² – 2AB cos(Z)

(08-03-2018 05:59 PM)Albert Chan Wrote: [ -> ]So, for right angle Z, A^2 + B^2 = C^2, as expected

That's also true for the correct formula. ;-)

Regarding the screenshot on the linked site:

The formulas refer both to the law of cosines (the three equations at the top) and the law of sines (smaller, right of the triangle), but here sides and angles have been swapped. It should read sin(X)/A = sin(Y)/B = sin(Z)/C.

So this looks like an error.

Dieter

My head get the correct formula, but when I type, somehow I switched the order ...

The screen shot was right in front of me ! Sorry ...

It should be this:

C^2 = A^2 + B^2 - 2 A B cos(Z)

So, if Z is right triangle, we get C^2 = A^2 + B^2

If Z is tiny, cos(Z) ~ 1,

C^2 ~ A^2 + B^2 - 2 A B ~ (A - B)^2

C ~ abs(A - B)

If Z is almost 180 degree, coz(Z) ~ -1:

C^2 ~ A^2 + B^2 + 2 A B ~ (A + B)^2

C ~ A + B

Dieter just answer your question earlier ...

(08-03-2018 06:08 PM)Dieter Wrote: [ -> ]It should read sin(X)/A = sin(Y)/B = sin(Z)/C.

While we are on the subject on triangle, Law of Cosine formula is not suited for needle-like triangle.

Prof. Kahan had the answer (see example 5) :

Miscalculating Area and Angles of a Needle-like Triangle
c² = a² + b² - 2ab*cosC =

(a-b)² + 4ab*(sin(C/2))² = (a-b)² + 4Δ tan(C/2)
Triangle area, Δ = ½ ab sinC

Thank you Albert, Dieter!