If MN=NQ and MQ=QR=RP, calculate for x​

C. 18°

Step-by-step explanation:

Refer to attached diagramGiven MN = NQ MQ = QR = RPNMR = 3xQMR = xTo findValue of xSolutionMN = NQ  ⇒ ∠MRN = ∠QMN = 3x+x = 4xMQ = QR ⇒ ∠MRQ = ∠QMR = x

RQP is exterior angle of ΔMQR ⇒

∠RQP = x + x = 2x

QR = RP ⇒

∠RPQ = ∠RQP = 2x

∠QRN is exterior angle of ΔQRP ⇒

∠QRN = 2x + 2x = 4x∠MRN = ∠QRN - ∠QRM = 4x - x = 3x

∠MRN = ∠NMR = 3x ⇒

NR = MQ

NR = MQ ⇒

∠NQR = ∠QRN = 4x

We now have a straight angle MQP:

∠MQP = ∠MQN + ∠RQN + ∠RQP180° = 4x + 4x + 2x10x = 180°x = 18°

Correct choice is C

X> -2 is the simplified version.