Question

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

Answer 1

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°