two circles touch eachother externally at c and ab is common tangents to the circles then angle acb is equal to?
Answers
Step-by-step explanation:
Given X and Y are two circles touch each other externally at C. AB is the common tangent to the circles X and Y at point A and B respectively.
To find : ∠ACB
Proof:
Let P be a point on AB such that, PC is at right angles to the Line Joining the centers of the circles.
Note that, PC is a common tangent to both circles.
This is because tangent is perpendicular to radius at point of contact for any circle.
let ∠PAC= α and ∠PBC = β.
PA = PC [lengths of the tangents from an external point C]
In a triangle CAP, ∠PAC = ∠ACP = α
similarly PB = CP and ∠PCB = ∠CBP = β
now in the triangle ACB,
∠CAB + ∠CBA + ∠ACB = 180° [sum of the interior angles in a triangle]
α + β + (α + β) = 180° (Since ∠ACB = ∠ACP + ∠PCB = α + β.
2α + 2β = 180°
α + β = 90°
∴ ∠ACB = α + β = 90°
the answer will be option D
solution
here is your answer hope its helpful
Answer:
90°
Step-by-step explanation:
Given X and Y are two circles touch each other externally at C. AB is the common tangent to the circles X and Y at point A and B respectively.
To find : ∠ACB
Proof:
Let P be a point on AB such that, PC is at right angles to the Line Joining the centers of the circles.
Note that, PC is a common tangent to both circles.
This is because tangent is perpendicular to radius at point of contact for any circle.
let ∠PAC= α and ∠PBC = β.
PA = PC [lengths of the tangents from an external point C]
In a triangle CAP, ∠PAC = ∠ACP = α
similarly PB = CP and ∠PCB = ∠CBP = β
now in the triangle ACB,
∠CAB + ∠CBA + ∠ACB = 180° [sum of the interior angles in a triangle]
α + β + (α + β) = 180° (Since ∠ACB = ∠ACP + ∠PCB = α + β.
2α + 2β = 180°
α + β = 90°
∴ ∠ACB = α + β = 90°