Step-by-step explanation:

...........................

Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Let assume that

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: \angle OBC=x \qquad \: \\ \\& \qquad \:\sf \: \angle BAC=y \end{aligned}} \qquad \: \\ \\ [/tex]

As we know, radius and tangent are perpendicular to each other.

Now, from the given figure, OB is radius and AB is tangent.

[tex]\sf\implies \sf \: OB \: \perp \: AB \\ \\ [/tex]

[tex]\sf\implies \sf \: \angle OBA \: = \: {90}^{ \circ} \\ \\ [/tex]

[tex]\sf \: \angle OBC + \angle ABC = {90}^{ \circ} \\ \\ [/tex]

[tex]\sf \: x + \angle ABC = {90}^{ \circ} \\ \\ [/tex]

[tex]\sf\implies \sf \:\angle ABC = {90}^{ \circ} - x - - - (1) \\ \\ [/tex]

Further we know that, length of tangents drawn from external point are equal.

As AB and AC are tangents drawn from external point A.

[tex]\sf\implies \sf \: AB = AC \\ \\ [/tex]

[tex]\sf\implies \sf \: \angle ACB = \angle ABC \\ \\ [/tex]

[tex]\sf\implies \sf \: \angle ACB = \angle ABC = {90}^{ \circ} - x\\ \\ [/tex]

Now, In triangle ABC,

We know, sum of interior angles of a triangle is 180°.

[tex]\sf \: \angle BAC + \angle ABC + \angle ACB = {180}^{ \circ} \\ \\ [/tex]

[tex]\sf \: y + {90}^{ \circ} - x + {90}^{ \circ} - x = {180}^{ \circ} \\ \\ [/tex]

[tex]\sf \: y + {180}^{ \circ} - 2x = {180}^{ \circ} \\ \\ [/tex]

[tex]\sf \: y - 2x = 0 \\ \\ [/tex]

[tex]\sf\implies \sf \: y = 2x \\ \\ [/tex]

[tex]\sf\implies \sf \: \angle BAC = 2\angle OBC\\ \\ [/tex]