SOLUTION: 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-
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-> SOLUTION: 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-
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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? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! This needs to be drawn:
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B
AC is 1/2 of AB in a 30-60-90 triangle.
AC is 5 units long.
Furthermore, slope of AC is negative reciprocal of BC.
Let C=(x,y)
slope of AC is (y-5)/(x-2)
slope of BC is (y+5)/(x-2)
Their product is -1
(y^2-25)/(x-2)^2 = -1
y^2-25=-x^2+4x-4
y^2=-x^2+4x+21
AC distance =5
square that to 25, and that equals (y-5)^2+(x-2)^2 (distance formula)
y^2-10y+25+x^2-4x+4=25
y^2-10y+x^2-4x-4=0
But y^2=-x^2+4x+21
-x^2+4x+21-10y+x^2-4x+4=0; the x^2 and the 4x cancel.
-10y+25=0
-10y=-25
y=2.5 THE Y-COORDINATE OF C.
25=(y-5)^2+(x-2)^2
25=6.25+(x-2)^2
18.75=(x-2)^2
x-2=4.33
x=6.33
