document.write( "Question 1143417: SinB=-1/2, 3pi/2 < B < 2pi; sinC= 1/4, pi/2 < C < pi find tan(B+C) \n" ); document.write( "
Algebra.Com's Answer #764241 by Edwin McCravy(20060)\"\" \"About 
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SinB=-1/2, 3pi/2 < B < 2pi; sinC= 1/4, pi/2 < C < pi find tan(B+C) \r
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document.write( "\"matrix%281%2C5%2Csin%28B%29=-1%2F2%2C+%22%22%2C%22%2C%22%2C%22%22%2C3pi%2F2+%3C+B+%3C+2pi%29\", \"matrix%281%2C5%2Csin%28C%29=1%2F4%2C+%22%22%2C%22%2C%22%2C%22%22%2Cpi%2F2+%3C+C+%3C+pi%29\" \r\n" );
document.write( "\"matrix%281%2C2%2Cfind%2C+tan%28B%2BC%29%29\" \r\n" );
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document.write( "First we draw angles B and C is standard position.\r\n" );
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document.write( "3pi/2 < B < 2pi means QIV. So we draw an angle with terminal side \r\n" );
document.write( "in QIV and assume the angle indicated by the red arc is theta:\r\n" );
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document.write( "Then we draw a perpendicular (in green) up to the x-axis from the \r\n" );
document.write( "end of the terminal side of B.  That makes a right triangle with \r\n" );
document.write( "the x-axis.\r\n" );
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document.write( "We know that the sine of B is -1/2.  We also know that the sine is \r\n" );
document.write( "the opposite over the hypotenuse, so we make the opposite side (the \r\n" );
document.write( "green side) equal to the numerator of -1/2, which is -1, (negative, \r\n" );
document.write( "since it goes downward). And we make the hypotenuse (the terminal\r\n" );
document.write( "side of B) be 2 (positive because the terminal side is always taken \r\n" );
document.write( "positive).  Then we calculate the adjacent side on the x-axis by the \r\n" );
document.write( "Pythagorean theorem:\r\n" );
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document.write( "\"c%5E2=a%5E2%2Bb%5E2\"\r\n" );
document.write( "\"2%5E2=a%5E2%2B1%5E2\"\r\n" );
document.write( "\"4=a%5E2%2B1\"\r\n" );
document.write( "\"3=a%5E2\"\r\n" );
document.write( "\"%22%22%2B-sqrt%283%29=a\"\r\n" );
document.write( "And since the adjacent side goes to the right of the origin we take\r\n" );
document.write( "the positive square root,\r\n" );
document.write( "\"sqrt%283%29=a\"\r\n" );
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document.write( "--------\r\n" );
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document.write( "pi/2 < C < pi means QII. So we draw an angle with terminal side \r\n" );
document.write( "in QII and assume the angle indicated by the red arc is C:\r\n" );
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document.write( " \r\n" );
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document.write( "Then we draw a perpendicular (in green) down to the x-axis from the \r\n" );
document.write( "end of the terminal side of C.  That makes a right triangle with \r\n" );
document.write( "the x-axis.\r\n" );
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document.write( "We know that the sine of B is 1/4.  We also know that the sine is the \r\n" );
document.write( "opposite over the hypotenuse, so we make the opposite side (the green \r\n" );
document.write( "side) equal to the numerator of 1/4, which is 1, (positive, since it \r\n" );
document.write( "goes up). And we make the hypotenuse (the terminal side of B) be 4 \r\n" );
document.write( "(since the terminal side is always taken positive).  Then we calculate \r\n" );
document.write( "the 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( "\"4%5E2=a%5E2%2B1%5E2\"\r\n" );
document.write( "\"16=a%5E2%2B1\"\r\n" );
document.write( "\"15=a%5E2\"\r\n" );
document.write( "\"%22%22%2B-sqrt%2815%29=a\"\r\n" );
document.write( "And since the adjacent side goes to the left of the origin we take\r\n" );
document.write( "the negative square root,\r\n" );
document.write( "\"-sqrt%2815%29=a\"\r\n" );
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document.write( "Now we have both angles A and B, so now we can proceed to finding\r\n" );
document.write( "tan(A+B)\r\n" );
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document.write( "   \r\n" );
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document.write( "\"tan%28A%2BB%29=%28tan%28A%29%2Btan%28B%29%29%2F%281-tan%28A%29tan%28B%29%29%29\"\"tangent=opposite%2Fadjacent\"\r\n" );
document.write( "\"tan%28A%29=%28-1%29%2Fsqrt%283%29=-1%2Fsqrt%283%29\"\r\n" );
document.write( "\"tan%28B%29=1%2F%28-sqrt%2815%29%29=-1%2Fsqrt%2815%29\"\r\n" );
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document.write( "Multiply every term by \"sqrt%283%29sqrt%2815%29\"\r\n" );
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document.write( "\"%28-sqrt%2815%29-sqrt%283%29%29%2F%28sqrt%283%29sqrt%2815%29-1%29\"\r\n" );
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document.write( "Rationalize the denominator by multiplying by \"%28sqrt%283%29sqrt%2815%29%2B1%29%2F%28sqrt%283%29sqrt%2815%29%2B1%29\"\r\n" );
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document.write( "\"%28-4%284sqrt%283%29%2Bsqrt%2815%29%29%29%2F44=%28-%284sqrt%283%29%2Bsqrt%2815%29%29%29%2F11\"\r\n" );
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document.write( "\"-%284sqrt%283%29%2Bsqrt%2815%29%29%2F11\"\r\n" );
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document.write( "Edwin
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