document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1204578>Question 1204578</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Here's another trig problem:<BR>\n" );
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document.write( "Thanks! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #840920 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=840920\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "The original diagram<BR>\n" );
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document.write( "The goal is to find the length of segment BC.\r<BR>\n" );
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document.write( "Use your favorite paint program to highlight triangle ABD. <BR>\n" );
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document.write( "This is a right triangle because of the square angle marker at angle ADB.<BR>\n" );
document.write( "The reference angle is 8 degrees. <BR>\n" );
document.write( "Opposite the reference angle is AD = 10.<BR>\n" );
document.write( "Adjacent to the reference angle is BD.<BR>\n" );
document.write( "Let's use the tangent trig ratio to find side BD. \r<BR>\n" );
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document.write( "Focus on triangle ABD only.<BR>\n" );
document.write( "tan(angle) = opposite/adjacent<BR>\n" );
document.write( "tan(B) = AD/BD<BR>\n" );
document.write( "tan(8) = 10/BD<BR>\n" );
document.write( "BD*tan(8) = 10<BR>\n" );
document.write( "BD = 10/tan(8)<BR>\n" );
document.write( "BD = 71.153697223842<BR>\n" );
document.write( "BD = 71.153697<BR>\n" );
document.write( "This value is approximate.<BR>\n" );
document.write( "Please make sure that your calculator is set to degrees mode.\r<BR>\n" );
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document.write( "Now let's highlight triangle ADC.<BR>\n" );
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document.write( "We'll use tangent once again since triangle ADC is a right triangle.<BR>\n" );
document.write( "tan(angle) = opposite/adjacent<BR>\n" );
document.write( "tan(C) = AD/CD<BR>\n" );
document.write( "CD = AD/tan(C)<BR>\n" );
document.write( "CD = 10/tan(11)<BR>\n" );
document.write( "CD = 51.445540159703<BR>\n" );
document.write( "CD = 51.445540\r<BR>\n" );
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document.write( "The summary so far is that we found these approximate side lengths<BR>\n" );
document.write( "BD = 71.153697<BR>\n" );
document.write( "CD = 51.445540<BR>\n" );
document.write( "They will be used in the next part to find side BC.\r<BR>\n" );
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document.write( "The last thing to highlight is triangle BCD.<BR>\n" );
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document.write( "Use the Law of Cosines.<BR>\n" );
document.write( "(BC)^2 = (BD)^2 + (CD)^2 - 2*(BD)*(CD)*cos(angle BDC)<BR>\n" );
document.write( "(BC)^2 = (71.153697)^2 + (51.445540)^2 - 2*(71.153697)*(51.445540)*cos(85)<BR>\n" );
document.write( "(BC)^2 = 7071.417954<BR>\n" );
document.write( "BC = sqrt(7071.417954)<BR>\n" );
document.write( "BC = 84.09172<BR>\n" );
document.write( "BC = 84\r<BR>\n" );
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document.write( "I rounded the final answer to the nearest whole number since the 10 cm given to us is a whole number. <BR>\n" );
document.write( "However, round that approximate value to some other level of precision if your teacher requires it.<BR>\n" );
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