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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 16 "The Law of Sines" }}{PARA 0 "" 
0 "" {TEXT -1 71 "\nDiscussion of the law of sines and applications to
 solving triangles.\n" }}{PARA 0 "" 0 "" {TEXT 258 153 "[Directions : \+
Execute the Code Resource section first. Although there will be no out
put immediately, these definitions are used later in this worksheet.]
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 
8 " 0. Code" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart; " }}}
}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 21 " 1.  The Law of Sines" }}{PARA 
0 "" 0 "" {TEXT -1 219 "\nThe law of sines is a three-sided equation. \+
It equates the ratio of the sine of each angle to the length of its op
posite side for all the angles and sides of a triangle. It is usually \+
used as a set of three equations.\n" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "sin(A)/a = sin(B)/b : % = sin(C)/c;" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 68 "sin(A)/a = sin(B)/b  ;\nsin(B)/b = sin(C)
/c  ;\nsin(A)/a = sin(C)/c  ;" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}}
{SECT 0 {PARA 4 "" 0 "" {TEXT -1 57 " 2.  Solving Triangles - Two Side
s & Included Angle Known" }}{PARA 0 "" 0 "" {TEXT -1 315 "\nThe law of
 cosines has four unknowns - the three sides and an angle. If we know \+
the three sides of a triangle, we can then find \"the angle\" - which \+
is the angle opposite of the \"hyptenuse\" (although this is not neces
sarily a right triangle and the side opposite of the angle is not nece
ssarily the longest side.)\n\n" }{TEXT 259 8 "        " }{TEXT 264 13 
"Example 2.1 :" }{TEXT -1 98 "   A triangle have sides of length : a =
 31,  b = 42, and angle A = 41 degrees. Find the angle B.\n" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "subs( \{a = 31, b = 42, A = 41*(Pi/
180) \}, sin(A)/a = sin(B)/b  );\nEq := evalf(%);" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 107 "sin(B) = solve( Eq, sin(B) );\nB = solve( E
q, B), ` radians`; \nB = evalf( solve( Eq, B)*180/Pi), ` degrees`;" }}
}{PARA 0 "" 0 "" {TEXT -1 3 "\n\n\n" }{TEXT 262 8 "        " }{TEXT 
265 13 "Example 2.2 :" }{TEXT -1 207 "    A triangle have sides of len
gth : a = 102, b = 79, and A = 22 degrees. Find c.\n\n\nIn this proble
m, we need to two steps. First, find angle B, and use this to find ang
le C. Then use angle C to find side c." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 116 "subs( \{A = 22*(Pi/180), a = 102, b = 79 \}, sin(A)/
a = sin(B)/b  );\nevalf(%);\nangle_B := solve( %, B): %, ` radians`;" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "angle_C := subs( \{A = 22
*(Pi/180), B = angle_B\}, Pi - A - C);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 104 "subs( \{A = 22*(Pi/180), a= 102, C = angle_C \}, sin
(A)/a = sin(C)/c  );\nevalf(%);\nside_c := solve( %, c);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT 
-1 57 " 3.  Solving Triangles - Two Angles & Included Side Known" }}
{PARA 0 "" 0 "" {TEXT -1 28 "\nHere is another scenario.\n\n" }{TEXT 
261 8 "        " }{TEXT 263 13 "Example 3.1 :" }{TEXT -1 345 "    A = \+
52 degrees,  B = 65 degrees, side c = 100. Find sides a and b.\n\nFirs
t we find angle C, which is easy to find. Then use one of the three eq
uations of the Law of Sines invovling A,a, C, and c. This is the basic
 method of using the Law of Sines - to find three of the four bits of \+
information about any two pairs, then solve for the fourth." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "C = 180 - 52 - 65;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "subs( \{A = 52*(Pi/180),  C = 63*(P
i/180), c = 100 \}, sin(A)/a = sin(C)/c  );\nEq := evalf(%);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "solve( %, a);\n" }}}{PARA 0 
"" 0 "" {TEXT -1 205 "\nNow that we know A, B, C, a, and c, we can fin
d b in a number of ways. Let's just find it the same way that we found
 a. (We could also relate A and B in the Law of Sines, or even use the
 Law of Cosines.)\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "subs(
 \{B = 65*(Pi/180),  C = 63*(Pi/180), c = 100 \}, sin(B)/b = sin(C)/c \+
 );\nEq := evalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "sol
ve( %, b);" }}}}{PARA 0 "" 0 "" {TEXT 260 50 "                \n      \+
  \251 2002 Waterloo Maple Inc" }}}{MARK "0 1" 35 }{VIEWOPTS 1 1 0 1 
1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }