| … | |
… | |
| 20 | DOCUMENT(); # This should be the first executable line in the problem. |
20 | DOCUMENT(); # This should be the first executable line in the problem. |
| 21 | |
21 | |
| 22 | loadMacros( |
22 | loadMacros( |
| 23 | "PG.pl", |
23 | "PG.pl", |
| 24 | "PGbasicmacros.pl", |
24 | "PGbasicmacros.pl", |
| 25 | "PGchoicemacros.pl", |
|
|
| 26 | "PGanswermacros.pl", |
25 | "PGanswermacros.pl", |
| 27 | "PGauxiliaryFunctions.pl" |
|
|
| 28 | ); |
26 | ); |
| 29 | |
27 | |
| 30 | TEXT(&beginproblem); |
28 | TEXT(&beginproblem); |
| 31 | $showPartialCorrectAnswers = 0; |
29 | $showPartialCorrectAnswers = 0; |
| 32 | |
30 | |
| 33 | $b=random(95,105,1); |
31 | $b=random(95,105); |
| 34 | |
32 | |
| 35 | TEXT(EV2(<<EOT)); |
33 | BEGIN_TEXT |
| 36 | A communication tower (the side CB) is located at the top (the point C) of |
34 | A communication tower (the side CB) is located at the top (the point C) of |
| 37 | a steep hill. |
35 | a steep hill. |
| 38 | The angle of inclination of the hill is 58 degrees. A guy wire is to be |
36 | The angle of inclination of the hill is 58 degrees. A guy wire is to be |
| 39 | attached to the top (the point B) of the tower and to the ground (the point A), |
37 | attached to the top (the point B) of the tower and to the ground (the point A), |
| 40 | $b m downhill from the base of the tower (the side AC). The angle |
38 | $b meters downhill from the base of the tower (the side AC). The angle |
| 41 | \( \angle BAC \) in the figure is 12 degrees. |
39 | \( \angle BAC \) in the figure is 12 degrees. |
| 42 | See the graph |
40 | See the graph |
|
|
41 | $BR$BR |
|
|
42 | $BCENTER |
| 43 | $PAR \{ image("c6s4p27.gif",tex_size=>200) \} $PAR |
43 | \{ image("c6s4p27.gif",tex_size=>200) \} $BR |
| 44 | $BBOLD Click on the graph to view a larger graph$EBOLD |
44 | $BBOLD Note: $EBOLD Click on the graph to view a larger graph |
|
|
45 | $ECENTER |
|
|
46 | $BR |
| 45 | $BR |
47 | $BR |
| 46 | Find the length of cable (the side AB) required for the guy wire. |
48 | Find the length of cable (the side AB) required for the guy wire. |
| 47 | Your answer is \{ans_rule(15)\} m; |
49 | $BR$BR |
| 48 | $BR |
50 | Answer: \{ans_rule(45)\} meters |
| 49 | EOT |
51 | END_TEXT |
| 50 | |
52 | |
| 51 | $ac=180-(90-58); |
53 | $ac=180-(90-58); |
| 52 | $ab=180-$ac-12; |
54 | $ab=180-$ac-12; |
| 53 | $ans1=$b*sin($PI*$ac/180)/sin($PI*$ab/180); |
|
|
| 54 | |
55 | |
| 55 | &ANS(std_num_cmp($ans1)); |
56 | ANS(num_cmp("$b*sin($ac pi/180)/sin($ab pi/180)")); |
| 56 | |
57 | |
| 57 | ENDDOCUMENT(); # This should be the last executable line in the problem. |
58 | ENDDOCUMENT(); # This should be the last executable line in the problem. |
Parent Directory
| 
Revision Log
| 
Patch