document.write( "Question 1013692: Triangle ABC is a 30˚ −60˚ −90˚ triangle with vertices A(2, 5) and B(2, −5). If m ∠B = 30˚ and ∠C is a right angle, what is the y-coordinate of C? \n" ); document.write( "
Algebra.Com's Answer #629993 by Boreal(15235)\"\" \"About 
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\n" ); document.write( "A
\n" ); document.write( ";;;;;;;;;;C
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\n" ); document.write( "B
\n" ); document.write( "AC is 1/2 of AB in a 30-60-90 triangle.
\n" ); document.write( "AC is 5 units long.
\n" ); document.write( "Furthermore, slope of AC is negative reciprocal of BC.
\n" ); document.write( "Let C=(x,y)
\n" ); document.write( "slope of AC is (y-5)/(x-2)
\n" ); document.write( "slope of BC is (y+5)/(x-2)
\n" ); document.write( "Their product is -1
\n" ); document.write( "(y^2-25)/(x-2)^2 = -1
\n" ); document.write( "y^2-25=-x^2+4x-4
\n" ); document.write( "y^2=-x^2+4x+21
\n" ); document.write( "AC distance =5
\n" ); document.write( "square that to 25, and that equals (y-5)^2+(x-2)^2 (distance formula)
\n" ); document.write( "y^2-10y+25+x^2-4x+4=25
\n" ); document.write( "y^2-10y+x^2-4x-4=0
\n" ); document.write( "But y^2=-x^2+4x+21
\n" ); document.write( "-x^2+4x+21-10y+x^2-4x+4=0; the x^2 and the 4x cancel.
\n" ); document.write( "-10y+25=0
\n" ); document.write( "-10y=-25
\n" ); document.write( "y=2.5 THE Y-COORDINATE OF C.
\n" ); document.write( "25=(y-5)^2+(x-2)^2
\n" ); document.write( "25=6.25+(x-2)^2
\n" ); document.write( "18.75=(x-2)^2
\n" ); document.write( "x-2=4.33
\n" ); document.write( "x=6.33\r
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