document.write( "Question 1141239: If sin(theta) = k and theta is obtuse, find an expression for tan (90 + theta) \n" ); document.write( "
Algebra.Com's Answer #761887 by Edwin McCravy(20060)\"\" \"About 
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document.write( "Obtuse means between 90° and 180°, exclusive of 90° and 180°.\r\n" );
document.write( "So theta is in QII. so we draw an angle with terminal side in QII \r\n" );
document.write( "and assume the angle indicated by the red arc is theta:\r\n" );
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document.write( "Then we drop a perpendicular (in green) down to the x-axis from the end of the\r\n" );
document.write( "terminal side of theta.  That makes a right triangle with the x-axis.\r\n" );
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document.write( "We know that the sine of theta is k, which is k/1.  We also know that\r\n" );
document.write( "the sine is the opposite over the hypotenuse, sow we make the opposite\r\n" );
document.write( "side (the green side) equal to the numerator of k/1, which is k, and we make the hypotenuse (the terminal side of theta) be 1.  Then we calculate the\r\n" );
document.write( "adjacent side on the x-axis by the Pythagorean theorem:\r\n" );
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document.write( "\"c%5E2=a%5E2%2Bb%5E2\"\r\n" );
document.write( "\"1%5E2=a%5E2%2Bk%5E2\"\r\n" );
document.write( "\"1=a%5E2%2Bk%5E2\"\r\n" );
document.write( "\"1-k%5E2=a%5E2\"\r\n" );
document.write( "\"%22%22%2B-sqrt%281-k%5E2%29=a\"\r\n" );
document.write( "And since the adjacent side goe to the left of the origin we take\r\n" );
document.write( "the negative square root,\r\n" );
document.write( "\"-sqrt%281-k%5E2%29=a\"\r\n" );
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document.write( "Now let's draw the angle 90°+theta by adding 90° to theta, indicated\r\n" );
document.write( "by the blue arrow, and we draw a perpendicular (in red) to the x-axis:\r\n" );
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document.write( "The lower triangle is congruent to the upper one.  Sides that go left\r\n" );
document.write( "or down are taken negative, so for 90°+theta we have:\r\n" );
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document.write( "Finally, since the tangent is the opposite over the adjacent, we have\r\n" );
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document.write( "Edwin
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