Answers

2016-03-12T10:53:42+05:30

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°

1 5 1
2016-03-12T11:00:44+05:30

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°


1 4 1
hey u cut and copy paste from Manishgoud035 ha ?
ya
ya
sry