In the given figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 54°, then ∠OBA = ____.

27°

It is given that tangents PA and PB are drawn from an external point P to a circle with centre O.

Therefore, PA = PB (Lengths of tangents drawn from an external point to a circle are equal)

In \(\Delta \)PAB,

PA + PB

Therefore, \(\angle\)PBA = \(\angle\)PAB  (in a triangle, equal sides have equal angles opposite to them)

Now,

\(\angle\)PAB + \(\angle\)PBA + \(\angle\)APB = \({180}^\circ\)  (Angle sum property)

⇒ 2\(\angle\)PAB + \({54}^\circ\) = \({180}^\circ\)

⇒ 2\(\angle\)PAB = \({180}^\circ\)  - \({54}^\circ\) = \({126}^\circ\)

⇒ \(\angle\)PAB = \({63}^\circ\)   ..(1)

Now, PA is the tangent and OA is the radius through the point of contact A.

Therefore, \(\angle\)OAP = \({90}^\circ\)  (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Now,

\(\angle\)OAB = \(\angle\)OAP - \(\angle\)PAB = \({90}^\circ\) - \({63}^\circ\) = \({27}^\circ\)  (using (1))

Therefore, \(\angle\)OAB is \({27}^\circ\).