Question

Two circles touch each other externally at P. AB is a common tangent to the circle touching them at A and Ben. find angle APR.

Answer 1

Hey there,

Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.

To find : ∠APB

Proof: let ∠CAP = α and ∠CBP = β.

CA = CP [lengths of the tangents from an external point C]

In a triangle PAC, ∠CAP = ∠APC = α

similarly CB = CP and ∠CPB = ∠PBC = β

now in the triangle APB,

∠PAB + ∠PBA + ∠APB = 180°   [sum of the interior angles in a triangle]

α + β + (α + β) = 180°

2α + 2β = 180°

α + β = 90°

∴ ∠APB = α + β = 90°

Hope this helps!

Answer 2

Answer:

Angle APB = 90 degree

Step-by-step explanation:

let, angle PAB = x

angle PBA = y

in triangle acp ,

ac=cp

so, angle apc = x

in triangle bpc,

pc = bc

so, angle bpc = y

in triangle APB

angle PAB +angle PBA + angle APB = 180 (angle sum property of triangle )

x + y + (x + y ) = 180

x + y +x + y =180

2 (x + y ) = 180

(x + y ) = 90

So, angle APB = 90 degree

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