Law of Sin by HP

08032018, 05:46 PM
Post: #1




Law of Sin by HP
Hi
On this HP application: https://www.hpcalc.org/details/4167 The diagram is a bit odd. Are not A, B and C supposed to be the angles; X, Y and Z the side lenghts? Adrian Coto 

08032018, 05:59 PM
(This post was last modified: 08112019 02:58 PM by Albert Chan.)
Post: #2




RE: Law of Sin by HP
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 

08032018, 06:08 PM
Post: #3




RE: Law of Sin by HP
(08032018 05:59 PM)Albert Chan Wrote: I just peek at your link, A, B, C are sides, X, Y, Z angles. I'd say A² + B² = C² + 2AB cos(Z) Or C² = A² + B² – 2AB cos(Z) (08032018 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 

08032018, 06:22 PM
Post: #4




RE: Law of Sin by HP
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 

08032018, 07:40 PM
Post: #5




RE: Law of Sin by HP
Comparing with Wolfram Alpha.
http://www.wolframalpha.com/input/?i=law+of+sines I think labels on HP diagram do not correspond to formulas, or viceversa. Adrian Coto 

08032018, 08:32 PM
(This post was last modified: 11112018 01:37 AM by Albert Chan.)
Post: #6




RE: Law of Sin by HP
Dieter just answer your question earlier ...
(08032018 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 needlelike triangle. Prof. Kahan had the answer (see example 5) : Miscalculating Area and Angles of a Needlelike Triangle c² = a² + b²  2ab*cosC = (ab)² + 4ab*(sin(C/2))² = (ab)² + 4Δ tan(C/2) Triangle area, Δ = ½ ab sinC 

08032018, 08:36 PM
Post: #7




RE: Law of Sin by HP
Thank you Albert, Dieter!
Adrian Coto 

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