A, B and C are three points on a circle with centre O. The tangent at C meets BA produced at T. If ∠ATC = 30° and ∠ACT = 48°,then what is the value of ∠AOB ?

Given : ∠ATC = 30° and ∠ACT = 48°

To find : ∠AOB = $$\theta$$ = ?

Solution : OA = OC = radius

=> $$\angle OAC=\angle OCA = x$$

Similarly, $$\angle OAB=\angle OBA = y$$ 

In $$\triangle$$ ACT,

=> $$\angle ACT+\angle ATC + \angle CAT=180^\circ$$

=> $$48^\circ+30^\circ \angle CAT=180^\circ$$

=> $$\angle CAT=180^\circ-78^\circ=102^\circ$$ -------------(i)

Also, radius intersects the tangent at the circumference of the circle at right angle.

=> $$\angle OCT=90^\circ$$

=> $$x+48^\circ=90^\circ$$

=> $$x=90^\circ-48^\circ=42^\circ$$ -------------(ii)

Now, at point A, => $$y+x+\angle CAT=180^\circ$$     [Supplementary angles]

Using equations (i) and (ii),

=> $$y+42^\circ+102=180^\circ$$

=> $$y=180^\circ-144^\circ=36^\circ$$ ------------(iii)

In $$\triangle$$ AOB,

=> $$\angle OBA+\angle OAB + \angle AOB=180^\circ$$

=> $$y+y+\theta=180^\circ$$

Using equation (iii), we get ;

=> $$(2 \times 36^\circ)+\theta=180^\circ$$

=> $$\theta=180^\circ-72^\circ=108^\circ$$

=> Ans - (D)