document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=828031>Question 828031</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Question: AT and BT are tangents to a circle, center O and radius 10cm. The length of the arc AB is 16cm. find (a)the size of angle AOB. (b)the area of triangle ABT. <BR>\n" );
document.write( "an image of the diagram i sketched <BR>\n" );
document.write( "http://postimg.org/image/eyky5f98h/\r<BR>\n" );
document.write( "\n" );
document.write( "i solved (a) and it was 1.6 radians, however i couldn't find a way to calculate the area of triangle ABT as the shape OATB is a Kite and we only have the radius, angle AOB and angles TAO=TBO=pi/2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #498978 by </B><A HREF=/tutors/aboutme.mpl?userid=KMST><B>KMST(5328)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=KMST><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/cgi-bin/show-face.mpl?user=KMST><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=498978\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> ABT is an isosceles triangle, with AT=BT.<BR>\n" );
document.write( "You could calculate its area as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28AT%29%2A%28BT%29%2Asin%28ATB%29%2F2=%28AT%29%5E2%2Asin%28ATB%29%2F2 ALIGN=MIDDLE  ALT=\"%28AT%29%2A%28BT%29%2Asin%28ATB%29%2F2=%28AT%29%5E2%2Asin%28ATB%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "You need the measure of angle ATB, and the length of side AT.<BR>\n" );
document.write( " <BR>\n" );
document.write( "Since you know 3 of the 4 angles in kite ATBO, you can easily calculate the measure of angle ATB, as<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2pi-pi%2F2-pi%2F2-1.6=pi-1.6=1.5416 ALIGN=MIDDLE  ALT=\"2pi-pi%2F2-pi%2F2-1.6=pi-1.6=1.5416\"   DISABLED_style= \"max-width:90%; height:auto;\" > (rounded)<BR>\n" );
document.write( " <BR>\n" );
document.write( "You can calculate AT from right triangle OAT.<BR>\n" );
document.write( "Angle AOT is half of angle AOB, and<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AT%2FAO=tan%28AOT%29 ALIGN=MIDDLE  ALT=\"AT%2FAO=tan%28AOT%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > so <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AT=AT%2Atan%28AOT%29=%2810cm%29%2Atan%280.8%29=10.3cm ALIGN=MIDDLE  ALT=\"AT=AT%2Atan%28AOT%29=%2810cm%29%2Atan%280.8%29=10.3cm\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( " <BR>\n" );
document.write( "So the area of ABT is<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%2810.3cm%29%5E2%2Asin%281.5416%29%2F2=53.0 ALIGN=MIDDLE  ALT=\"%2810.3cm%29%5E2%2Asin%281.5416%29%2F2=53.0\"   DISABLED_style= \"max-width:90%; height:auto;\" > (rounded).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );