document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1202951>Question 1202951</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Evaluate the expression:<BR>\n" );
document.write( "sin^(-1)⁡(2/5)+sin^(-1)(√21/5) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #838165 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13203)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=838165\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Because the inverse sine is positive for both angles, they are both in quadrant I.<br><BR>\n" );
document.write( "Draw a right triangle in standard position with a sine (opposite over hypotenuse) of 2/5 and use <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%5E2%28x%29%2Bcos%5E2%28x%29=1 ALIGN=MIDDLE  ALT=\"sin%5E2%28x%29%2Bcos%5E2%28x%29=1\"   DISABLED_style= \"max-width:90%; height:auto;\" > to determine that the cosine of the angle is sqrt(21)/5, which is the sine of the other angle in the given expression.<br><BR>\n" );
document.write( "In a right triangle, the sine of one acute angle is the cosine of the other, so the two angles in the given expression are the two angles in a right triangle.  So the sum of the two angles is 90 degrees.<br><BR>\n" );
document.write( "ANSWER: 90 degrees<br><BR>\n" );
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