Math, asked by rudeawakening, 1 year ago

XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. prove that


rudeawakening: <AOB = 90 deegrees

Answers

Answered by shadataj
903
I hope u got it ... plzz like
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rudeawakening: thanks for answering even though it's reported !!
Mohit0: but how cant both the radii r straight u cant take linear pair in that que. if answer on this but unluckuly i cant post it
rudeawakening: i didn't understood ?
rudeawakening: will you pl tell again?
Mohit0: she has taken op and oq radius to be a straight line it cant be
rudeawakening: u mean she has skiped a step
Mohit0: no bro she has considered both the radii to be straight that is they make an angle of 180 betn them but the two radii dont make straight line
rudeawakening: bt if the tangents are parallel then it will be in straight line na?
Mohit0: o ya thnks to correct me
rudeawakening: your welcome !!
Answered by karishmad
290
GIVEN- a circle with centre o to which XY and XY' are tangents
TP- ∠AOB=90°
CONST.- join OY, OY' and OC
PROOF- In ΔOYA and ΔOCA
             OY = OC [radii]
             OA=OA [common]
            AY = AC [tangents]
⇒By SSS
    ΔOYA≈ΔOCA
⇒∠OY'A=∠OCA   [CPCT]---------------(1)
║ly ΔOY'B≈ΔOCB
⇒∠Y'BO=∠CBO   [CPCT]--------------------(2)
∠YAB+∠Y'BA=180°[co-interior angles]
2∠OAB+2∠OBA=180°      -------[from (1) and (2)]
2(∠OAB+∠OBA)=180°
∠OAB+∠OBA=90°---------------------(3)
In ΔAOB
∠OAB+∠OBA+∠AOB=180°   [Angle sum property of a triangle]
90°+∠AOB=180°
∠AOB=180°- 90°
∠AOB=90°
∴ Hence Proved


rudeawakening: thanks
karishmad: your welcome
karishmad: all the best for tommorow
rudeawakening: same to you!!
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