Question

Two circles touch each other externally at P. AB is a common tangent to the circle touching them at A and B . Find angle APB

Answer 1

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°



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Answer 2

To find :-

$\angle{APB}= ?$

Proof:-

$\small\sf{\angle{CAP}+\angle{APC+}\angle{CPB}+\angle{BCP}+=180°}$

$\small\sf{\angle{x}+\angle{x}+\angle{y}+\angle{y}=180°}$

$\small\sf{2\angle{x}+2\angle{y}=180°}$

$\small\sf{x+y=\frac{180}{2}}$

$\small\sf{x+y=90}$

$\small\sf{\angle{APB}=90°}$