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XY bisects ∠AXB. If m∠AXB = -18x° and m∠AXY = (3x2 - 12)°. Find m∠BXY. Please help, I'm blanking!!!

Sagot :

abidemiokin

Answer:

m<BXY = 36degrees

Step-by-step explanation:

If XY bisects ∠AXB, then;

<AXY + <BXY = <AXB

Given

m∠AXY = (3x^2 - 12)°

m∠AXB = -18x°

Required

Find m∠BXY.

From the formula above;

Find m∠BXY = <AXB - <AXY

m<BXY = -18x - (3x^2-12)

m<BXY = -18x - 3x^2 + 12

m<BXY =  -3x^2 -18x + 12

Also <AXY = m<BXY

3x^2 - 12 =   -3x^2 -18x + 12

6x^2 + 18x -24 = 0

x^2+3x-4 = 0

Factorize

x = -3±√9+16/2

x = -3±5/2

x = -3+5/2 and -3-5/2

x = 2/2 and -8/2

x = 1 and -4

Substitute x = 1 into m<BXY

m<BXY =  -3x^2 -18x + 12

m<BXY =  -3(1)^2 -18(1) + 12

m<BXY =  -3 -18+ 12

m<BXY =  -9

when x= -4

m<BXY =  -3(-4)^2 -18(-4) + 12

m<BXY =  -3(16) +72+ 12

m<BXY =  -48+84

m<BXY = 36degrees

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