HP 20S and HP 21S: Triangle Program
Given three Cartesian
coordinates, through points (R1, R4), (R2, R5), and (R3, R6), this program
calculates:
1. The lengths
of each side, represented by R7, R8, and R9.
2. The angle
between lines connected by (R1, R4)-(R2, R5) and (R2, R5)-(R3, R6). The angle is stored in R0.
3. The area of
the triangle. This is the final results
shown.
The keystrokes
for the HP 20S and HP 21S in this program are the same. Store the coordinates in variables R1 through
R6, then press XEQ A.
HP 20S and HP 21S Program:
Triangle Program
STEP
|
CODE
|
KEY
|
01
|
61, 41, A
|
LBL A
|
02
|
22, 1
|
RCL 1
|
03
|
65
|
-
|
04
|
22, 2
|
RCL 2
|
05
|
31
|
INPUT
|
06
|
22, 4
|
RCL 4
|
07
|
65
|
-
|
08
|
22, 5
|
RCL 5
|
09
|
51, 21
|
>P
|
10
|
51, 31
|
SWAP
|
12
|
26
|
R/S
|
13
|
71
|
[ C ] Clear
|
14
|
22, 2
|
RCL 2
|
15
|
65
|
-
|
16
|
22, 3
|
RCL 3
|
17
|
31
|
INPUT
|
18
|
22, 5
|
RCL 5
|
19
|
65
|
-
|
20
|
22, 6
|
RCL 6
|
21
|
51, 21
|
>P
|
22
|
51, 31
|
SWAP
|
23
|
21, 8
|
STO 8
|
24
|
26
|
R/S
|
25
|
71
|
[ C ] (Clear)
|
26
|
22, 1
|
RCL 1
|
27
|
65
|
-
|
28
|
22, 3
|
RCL 3
|
29
|
31
|
INPUT
|
30
|
22, 4
|
RCL 4
|
31
|
65
|
-
|
32
|
22, 6
|
RCL 6
|
33
|
51, 21
|
>P
|
34
|
51, 31
|
SWAP
|
35
|
21, 9
|
STO 9
|
36
|
26
|
R/S
|
37
|
71
|
[ C ] Clear
|
38
|
22, 8
|
RCL 8
|
39
|
51, 11
|
x^2
|
40
|
75
|
+
|
41
|
22, 9
|
RCL 9
|
42
|
51, 11
|
x^2
|
43
|
65
|
-
|
44
|
22, 7
|
RCL 7
|
45
|
51, 11
|
x^2
|
46
|
74
|
=
|
47
|
45
|
÷
|
48
|
2
|
2
|
49
|
45
|
÷
|
50
|
22, 8
|
RCL 8
|
51
|
45
|
÷
|
52
|
22, 9
|
RCL 9
|
53
|
74
|
=
|
54
|
51, 24
|
ACOS
|
55
|
21, 0
|
STO 0
|
56
|
26
|
R/S
|
57
|
23
|
SIN
|
58
|
55
|
*
|
59
|
22, 7
|
RCL 7
|
60
|
55
|
*
|
61
|
22, 8
|
RCL 8
|
62
|
45
|
÷
|
63
|
2
|
2
|
64
|
74
|
=
|
65
|
61, 26
|
RTN
|
Example:
Inputs:
(R1, R4) = (4,
8)
(R2, R5) = (2,
3)
(R3, R6) = (3,
10)
Outputs:
R7 = 5.3852
R8 = 7.0711
R9 = 2.2361
R0 = 34.6952°
Area = 10.8374
Eddie
This blog is
property of Edward Shore, 2017