# 2 circles touch each other externally at p .ab is a common tangent to the circles touching them at a and b .the value of angle apb is

1
by amanthakkar

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by amanthakkar

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Now,

NA = NP and NP = NB (tangents drawn from same point onto same circles)

∴ ΔNAP and ΔNBP are isosceles triangles.

Let ∠NAP = x° and ∠NBP = y°

In ΔNAP,

=> ∠NAP = ∠NPA = x° (base angles)

=> ∠NAP+∠NPA+∠PNA = 180°

=> ∠PNA = 180-x-x = (180-2x)°

Similarly, in ΔNBP,

=> ∠PNB = (180-2y)°

∠PNB+∠PNA = 180° (linear pairs)

=> (180-2x)°+(180-2y)° = 180°

=> 2y+2x = 180

=> x+y = 90° (dividing equation by 2) ... (1)

Now, in ΔAPB,

∠NAP+∠NBP+∠APB = 180°

=> x+y+∠APB = 180

=> 90+∠APB = 180 (from 1)

=> ∠APB = 180-90 =