Answers

2016-02-23T09:15:43+05:30
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 = 90°
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