(11C) Tree Heights
11-22-2018, 05:27 PM (This post was last modified: 11-22-2018 05:35 PM by ijabbott.)
Post: #11
 ijabbott Senior Member Posts: 625 Joined: Jul 2015
RE: (11C) Tree Heights
(11-04-2018 04:57 PM)Dieter Wrote:
(11-04-2018 03:32 PM)Thomas Klemm Wrote:  We don't really need trigonometric functions here.
Good old Pythagoras is good enough:

Great! This way it can also be done on the 12C and other calculators without trigs or polar/rectangular conversion:

Code:
01 X<>Y 02 - 03 LstX 04 ENTER 05 x 06 EEX 07 4 08 + 09 √x 10 ÷ 11 × 12 GTO 00

Since no →P is required this may even run slightly faster than Thomas' original version. If available, replace "ENTER x" with x².

(11-04-2018 03:32 PM)Thomas Klemm Wrote:  Examples:

56 ENTER
20 ENTER
40 A
10.9825

56 ENTER
-20 ENTER
40 A
32.9475

Same for the above version. Press [R/S] instead of [A]. ;-)

That's a neat solution! It's also worth mentioning that if you know the tangent, sine or cosine of an angle between 0 and 90 degrees, you can derive the others with standard arithmetic and the square root function.

$$\tan(x) = \frac{\sqrt{1-\cos^2(x)}}{\cos(x)}$$, or: $$\tan(x) = \sqrt{\frac{1}{\cos^2(x)} - 1}$$

Code:
ENTER × 1/x 1 - √x

$$\cos(x) = \frac{1}{\sqrt{1 + \tan^2(x)}}$$

Code:
ENTER × 1 + √x 1/x

$$\tan(x) = \frac{\sin(x)}{\sqrt{1 - \sin^2(x)}}$$

Code:
ENTER ENTER × 1 X<>Y - √x ÷

$$\sin(x) = \frac{\tan(x)}{\sqrt{1 + \tan^2(x)}}$$

Code:
ENTER ENTER × 1 + √x ÷

$$\sin(x) = \sqrt{1 - \cos^2(x)}$$, and: $$\cos(x) = \sqrt{1 - \sin^2(x)}$$

Code:
ENTER × 1 X<>Y - √x

Of course, "ENTER", "×" can be replaced by "x²" in all of the above, if available.

— Ian Abbott
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 Messages In This Thread (11C) Tree Heights - Gamo - 11-02-2018, 01:24 PM RE: (11C) Tree Heights - SlideRule - 11-02-2018, 02:43 PM RE: (11C) Tree Heights - Dieter - 11-02-2018, 06:15 PM RE: (11C) Tree Heights - SlideRule - 11-02-2018, 07:03 PM RE: (11C) Tree Heights - Dieter - 11-02-2018, 07:41 PM RE: (11C) Tree Heights - Gamo - 11-03-2018, 01:35 AM RE: (11C) Tree Heights - Dieter - 11-03-2018, 12:47 PM RE: (11C) Tree Heights - Thomas Klemm - 11-04-2018, 03:32 PM RE: (11C) Tree Heights - Dieter - 11-04-2018, 04:57 PM RE: (11C) Tree Heights - ijabbott - 11-22-2018 05:27 PM RE: (11C) Tree Heights - Gamo - 11-05-2018, 12:52 AM

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