In the given figure, BC is the diameter of the circle with the centre O and PAT is the tangent at A. If Ð ABC = 38^o, find Ð BAT. No figure though.

∠BAC = 90°  [Angle in a semi circle is right angle]

Also, ∠OAT = 90°  [Angle made by the radius and the tangent at the point of contact is 90°]

∴ ∠BAC = ∠OAT

⇒ ∠BAO + ∠OAC = ∠OAC + ∠CAT

⇒ ∠BAO = ∠CAT

In ∆AOB,

OA = OB = r (radius)

∴ ∆AOB is isosceles.

⇒ ∠OBA = ∠OAB = 38°

∴ ∠CAT = 38°

∠BAT = ∠BAC + ∠CAT

= 90° + 38°

= 128°