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PLEASE HELP GEOMETRY ANGLES!!

PLEASE HELP GEOMETRY ANGLES!!-example-1

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Answer:

∠PQS = ∠PQR+∠RQS

x²-5 = 45+5x

x²-5-45-5x=0

x²-5x-50=0

b²-4ac = (-5)²-4(1*-50)= 225

Δ=225; √Δ=15

Δ is strictly positive, the equation x²−5x−50=0 admit two solutions.

(-b-√Δ)/2a = (5-15)/2 = -10/2=-5

(-b+√Δ)/2a = (5+15)/2 =10

x = -5 or 10

a measurement is never negative, x= 10°

∠PQS = x²-5 = 10²-5 = 95°

we check :

∠PQS= (95°)=∠PQR+∠RQS = 45+5x = 45*5*10 =95°

User Ccellar
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