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{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 54 "High School Modul
es > Trigonometry by Gregory A. Moore" }{TEXT 266 1 " " }}{PARA 3 "" 
0 "" {TEXT -1 4 "    " }{TEXT 256 18 "The Law of Cosines" }}{PARA 0 "
" 0 "" {TEXT -1 73 "\nDiscussion of the law of cosines and application
s to solving triangles.\n" }}{PARA 0 "" 0 "" {TEXT 258 153 "[Direction
s : Execute the Code Resource section first. Although there will be no
 output immediately, these definitions are used later in this workshee
t.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT 
-1 8 " 0. Code" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart; w
ith(plots): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 526 "TriPlot :=
 proc(A,B,C)\nlocal sA,sB,sC,d;\nd := 1/4:\nsA := [ (C[1]+B[1])/2, (C[
2]+B[2])/2];\nsB := [ (A[1]+C[1])/2, (A[2]+C[2])/2];\nsC := [ (A[1]+B[
1])/2, (A[2]+B[2])/2];\ndisplay(\n    plot( [A,B,C,A], color = blue, t
hickness = 3, axes = none ),\n    textplot( [A[1]-d, A[2]-d, `angle A`
]),\n    textplot( [B[1]+d, B[2]+d, `angle B`]),\n    textplot( [C[1]+
d, C[2]-d, `angle C`]),\n    textplot( [sA[1]+2*d, sA[2], `side a`]),
\n    textplot( [sB[1], sB[2]+d, `side b`]),\n    textplot( [sC[1]-2*d
, sC[2]+d, `side c`])  \n        );\nend:\n" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 775 "TriPlotN := proc(A,B,C)\nlocal sA,sB,sC,d, da,db,
dc;\n\nd := 1/4:\nsA := [ (C[1]+B[1])/2, (C[2]+B[2])/2];\nsB := [ (A[1
]+C[1])/2, (A[2]+C[2])/2];\nsC := [ (A[1]+B[1])/2, (A[2]+B[2])/2];\nda
 := evalf(sqrt( (C[1]-B[1])^2  + (C[2]-B[2])^2 ),3);\ndb := evalf(sqrt
( (C[1]-A[1])^2  + (C[2]-A[2])^2 ),3);\ndc := evalf(sqrt( (A[1]-B[1])^
2  + (A[2]-B[2])^2 ),3);\ndisplay(\n    plot( [A,B,C,A], color = blue,
 thickness = 3, axes = none ),\n    textplot( [A[1]-d, A[2]-d, `angle \+
A`]),\n    textplot( [B[1]+d, B[2]+d, `angle B`]),\n    textplot( [C[1
]+d, C[2]-d, `angle C`]),\n    textplot( [sA[1]+3*d, sA[2],   cat(`a =
 `,convert(da,string))]),\n    textplot( [sB[1], sB[2]+d,     cat(`b =
 `,convert(db,string))]),\n    textplot( [sC[1]-2*d, sC[2]+d, cat(`c =
 `,convert(dc,string))])  \n        );\nend:\n" }}}}{SECT 0 {PARA 4 "
" 0 "" {TEXT -1 23 " 1.  The Law of Cosines" }}{PARA 0 "" 0 "" {TEXT 
-1 100 "\nThe law of cosines is a generalization of Pythagoras' theore
m that applies to non-right triangles.\n" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "LoC := a^2 + b^2 -2*a*b*cos(theta) = c^2;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "TriPlot([0,0],[10,6],[7,0]);" }}}
{EXCHG }{PARA 0 "" 0 "" {TEXT -1 163 "\nHere is a table showing how ch
anging the angle of a triangle affects the length of the opposite side
 (keeping the two sides which form the angle fixed in length)." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 259 "n := 8:    a:= 3: b := 4:\n
A := array( [seq( [ i*Pi/n, a, b, evalf(sqrt(a^2 + b^2 -2*a*b*cos(i*Pi
/n))) ],   \n            i = -1..n ) ]):\nA[1,1] :=theta: A[1,2]:= 'a'
: A[1,3] := 'b': A[1,4]:=`c`:\n`With a & b fixed, theta increasing mak
es c increase`;  print(A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
260 "n := 8:    a:= 5: b := 12:\nA := array( [seq( [ i*Pi/n, a, b, eva
lf(sqrt(a^2 + b^2 -2*a*b*cos(i*Pi/n))) ],   \n            i = -1..n ) \+
]):\nA[1,1] :=theta: A[1,2]:= 'a': A[1,3] := 'b': A[1,4]:=`c`:\n`With \+
a & b fixed, theta increasing makes c increase`;  print(A);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "a:= 'a': b:= 'b': c:= 'c':" 
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 
48 " 2.  Solving Triangles - Two Sides & Angle Known" }}{PARA 0 "" 0 "
" {TEXT -1 1 "\n" }{TEXT 259 21 "        Example 2.1 :" }{TEXT -1 82 "
  a = 100, b = 140, and the angle between them is 38 degrees. Find the
 third side." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "rad_angle :=
 38*(Pi/180);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "subs( \{a \+
= 100, b = 140, theta = rad_angle\}, LoC);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 9 "evalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
23 "solve(%, c):\nabs(%[1]);" }}}{PARA 0 "" 0 "" {TEXT -1 52 "\nOf cou
rse, only the positive solution makes sense.\n" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 73 "TriPlotN([0,0],[140 + 100*cos(142*Pi/180), 100*s
in(142*Pi/180)],[140,0]);" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n\n" }
{TEXT 264 21 "        Example 2.2 :" }{TEXT -1 173 "  A right triangle
 has sides 7, 24, and 25. If the two bases are fixed in length, and th
e right angle is \n        increased to 95 degrees, how long will the \+
hypotenuse be? \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "rad_ang
le := 95*(Pi/180);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "subs(
 \{a = 7, b = 24, theta = rad_angle\}, LoC);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 33 "evalf(%);\nsolve(%, c):\nabs(%[1]);" }}}{PARA 0 
"" 0 "" {TEXT -1 2 "\n\n" }{TEXT 265 20 "       Example 2.3 :" }{TEXT 
-1 189 "  A right triangle has sides 9, 40, and 41. If the two bases a
re fixed in length, and the right angle is \n        increased to 10%,
 what percentage increase will the hypotenuse experience?\n" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "rad_angle := (Pi/2)*1.10;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "subs( \{a = 9, b = 40, theta
 = rad_angle\}, LoC);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "ev
alf(%);\nsolve(%, c):\nabs(%[1]);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 5 "%/41;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "%-1
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "%*100,%;" }}}{PARA 0 "
" 0 "" {TEXT -1 2 "\n\n" }{TEXT 260 21 "        Example 2.4 :" }{TEXT 
-1 209 "  A bug walks from the origin along the x-axis to the point (2
00,0). Then the bug turns left, and heads \n        off at 71 degrees \+
above the axis, and walks 100 units. How far is the bug away from the \+
origin?\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "angle := 180 - \+
71;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "rad_angle := (Pi/2)*
angle;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "subs( \{a = 200, \+
b = 100, theta = rad_angle\}, LoC);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "evalf(%);\nsolve(%, c):\nabs(%[1]);" }}}}{SECT 0 
{PARA 4 "" 0 "" {TEXT -1 42 " 3.  Solving Triangles - Three Sides Know
n" }}{PARA 0 "" 0 "" {TEXT -1 315 "\nThe law of cosines has four unkno
wns - the three sides and an angle. If we know the three sides of a tr
iangle, we can then find \"the angle\" - which is the angle opposite o
f the \"hyptenuse\" (although this is not necessarily a right triangle
 and the side opposite of the angle is not necessarily the longest sid
e.)\n\n" }{TEXT 262 21 "        Example 3.1 :" }{TEXT -1 70 "     a = \+
31,  b = 42, c = 22. Find the angle formed by sides a and b.\n" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "subs( \{a = 31, b = 42, c = \+
22 \}, LoC);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "solve( %, t
heta):\n% = evalf(%), `radians`;\nevalf( %%*180/Pi), `degrees`;" }}}
{PARA 0 "" 0 "" {TEXT -1 204 "\nIn fact, a triangle has three angles a
nd any of them can be considered the angle in the formula. This is sel
ected implicitly by selecting which side is c - which is opposite of t
he angle in question.\n\n\n\n" }{TEXT 261 21 "        Example 3.2 :" }
{TEXT -1 49 "    a = 10, b = 9, c = 8. Find all three angles.\n" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "`the angle formed by sides a
 & b is opposite of side c`;\nsubs( \{a = 10, b = 9, c = 8\}, LoC);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "Angle_ab := solve( %, thet
a);\n% = evalf(%),` radians`; %% = evalf( %%*180/Pi), ` degrees`;" }}}
{PARA 0 "" 0 "" {TEXT -1 261 "\nOk, we found the angle formed by sides
 a & b. What about the other angles? We can do this is two different w
ays - relabel the sides (now we call \"a\" by \"c\" and \"c\" by \"a\"
 and use the same formula) or create a new version of the formula with
 a and c reversed. \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "`th
e angle formed by sides a & c is opposite of side b`;\nsubs( \{a = 8, \+
b = 9, c = 10\}, LoC);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "A
ngle_ac := solve( %, theta);\n% = evalf(%),` radians`; %% = evalf( %%*
180/Pi), ` degrees`;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "`t
he angle formed by sides a & c is opposite of side b`;\nsubs( \{a = 10
, b = 9, c = 8\}, c^2 + b^2 -2*c*b*cos(theta) = a^2);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "Angle_ac := solve( %, theta);\n% = \+
evalf(%),` radians`; %% = evalf( %%*180/Pi), ` degrees`;" }}}{PARA 0 "
" 0 "" {TEXT -1 77 "\nWe get the same answer either way of course. The
 third angle is easy to find" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
99 "Angle_bc := Pi - Angle_ab - Angle_ac; \n% = evalf(%),` radians`; %
% = evalf( %%*180/Pi), ` degrees`;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 86 "Angle_ab, Angle_bc, Angle_ac: \nevalf(%,3), ` radians
`; \nevalf(%%*180/Pi,3),` degrees`;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT 263 53 "                \n
        \251 2002 Waterloo Maple Inc   " }}}{MARK "0 1" 17 }{VIEWOPTS 
1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }