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Question-184041




Question Number 184041 by Acem last updated on 02/Jan/23
Answered by HeferH last updated on 02/Jan/23
TN^(⌢)  = (π/3) = ((180°)/3) = 60°  ⇒ ∠TAN = ((TN^(⌢) )/2) = 30°   ∠AOT = 120° (△AOT isosceles)    ∠ATC = ((AT^(⌢) )/2) = 60  ⇒ ∠ACT = 90°    MTCA is cyclic,   the center would be the mid point of AT
TN=π3=180°3=60°TAN=TN2=30°AOT=120°(AOTisosceles)ATC=AT2=60ACT=90°MTCAiscyclic,thecenterwouldbethemidpointofAT
Answered by mr W last updated on 02/Jan/23
Commented by mr W last updated on 02/Jan/23
∠TAN=((TN^(⌢) )/2)=30°  ∠BOT=2∠OAT=60°=∠OAC  ⇒AC//OT  ⇒AC⊥BC  ⇒∠ACB=90°=∠AMT  ⇒ACTM are cyclic, lie on a circle       with AT as diameter. Midpoint       of AT is the center.
TAN=TN2=30°BOT=2OAT=60°=OACAC//OTACBCACB=90°=AMTACTMarecyclic,lieonacirclewithATasdiameter.MidpointofATisthecenter.

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