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    1 ##DESCRIPTION
    2 ## Algebra: Trigonometry of right triangles
    3 ##ENDDESCRIPTION
    4 
    5 ##KEYWORDS('algebra', 'trigonometry', 'The law of sines')
    6 
    7 ## tcao , PAID on 11-24-2003
    8 
    9 ## DBsubject('Trigonometry')
   10 ## DBchapter('Trigonometric Functions of Angles')
   11 ## DBsection('The Law of Sines')
   12 ## Date('6/3/2002')
   13 ## Author('')
   14 ## Institution('')
   15 ## TitleText1('Precalculus')
   16 ## EditionText1('3')
   17 ## AuthorText1('Stewart, Redlin, Watson')
   18 ## Section1('6.4')
   19 ## Problem1('27')
   20 DOCUMENT();        # This should be the first executable line in the problem.
   21 
   22 loadMacros(
   23 "PG.pl",
   24 "PGbasicmacros.pl",
   25 "PGchoicemacros.pl",
   26 "PGanswermacros.pl",
   27 "PGauxiliaryFunctions.pl"
   28 );
   29 
   30 TEXT(beginproblem());
   31 $showPartialCorrectAnswers = 0;
   32 
   33 $b=random(95,105,1);
   34 
   35 TEXT(EV2(<<EOT));
   36 A communication tower (the side CB) is located at the top (the point C) of
   37 a steep hill.
   38 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),
   40 $b m downhill from the base of the tower (the side AC). The angle
   41 \( \angle BAC \) in the figure is 12 degrees.
   42 See the graph
   43 $PAR \{ image("c6s4p27.gif") \} $PAR
   44 $BBOLD Click on the graph to view a larger graph$EBOLD
   45 $BR
   46 Find the length of cable (the side AB) required for the guy wire.
   47 Your answer is \{ans_rule(15)\} m;
   48 $BR
   49 EOT
   50 
   51 $ac=180-(90-58);
   52 $ab=180-$ac-12;
   53 $ans1=$b*sin($PI*$ac/180)/sin($PI*$ab/180);
   54 
   55 ANS(num_cmp($ans1));
   56 
   57 ENDDOCUMENT();        # This should be the last executable line in the problem.