Question

find the value of x in the question

Answer 1

Given : QT bisects PR, QS = PS, ∠TQR = 40, ∠RPS = 20

To find x:

name the point 0 at which the line PS and QT intersect

∠POT + 20 + 90 = 180 ( angle sum property )

∠POT = 180- 110

          = 70

∠POT = ∠SOQ (V.O.A)

∠OSQ + ∠SOQ + 40 = 180 ( angle sum property)

∠OSQ + 70 + 40 = 180

∠OSQ = 180 - 110

           = 70

∵ PQ║PS,

∴ ∠PQS = PSQ

x + 40 = 70 ( substituting )

x = 70 - 40

x = 30

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