Law of Sin by HP
08-03-2018, 05:46 PM
Post: #1 acoto Junior Member Posts: 33 Joined: Sep 2017
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?

08-03-2018, 05:59 PM (This post was last modified: 08-11-2019 02:58 PM by Albert Chan.)
Post: #2
 Albert Chan Senior Member Posts: 1,598 Joined: Jul 2018
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
08-03-2018, 06:08 PM
Post: #3
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: Law of Sin by HP
(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
08-03-2018, 06:22 PM
Post: #4
 Albert Chan Senior Member Posts: 1,598 Joined: Jul 2018
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
08-03-2018, 07:40 PM
Post: #5 acoto Junior Member Posts: 33 Joined: Sep 2017
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.

08-03-2018, 08:32 PM (This post was last modified: 11-11-2018 01:37 AM by Albert Chan.)
Post: #6
 Albert Chan Senior Member Posts: 1,598 Joined: Jul 2018
RE: Law of Sin by HP

(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
08-03-2018, 08:36 PM
Post: #7 acoto Junior Member Posts: 33 Joined: Sep 2017
RE: Law of Sin by HP
Thank you Albert, Dieter!