In ATC, the sum of all interior angles is 180^°. Given that = 30^° and = 48^°, we find = 180^° - 78^° = 102^°.

Points T, A, and B lie on a straight line because side BA is produced to T. Hence, and form a linear pair, giving = 180^° - 102^° = 78^°. Alternatively, is an exterior angle to ABC, so = + = 102^°.

Points T, A, and B lie on a straight line because side BA is produced to T. Hence, and form a linear pair, giving = 180^° - 102^° = 78^°. Alternatively, is an exterior angle to ABC, so = + = 102^°.

In ABC, the sum of angles is 180^°. Using the values = 78^° and = 48^°, we calculate = 180^° - 126^° = 54^°.

43. $\text{A}$ , $\text{B}$ और $\text{C}$ केन्द्र $\text{O}$ वाले वृत्त पर स्थित तीन बिन्दु हैं। स्पर्श रेखा $\text{C}$ , $\text{BA}$ को $\text{T}$ तक बढ़ाए जाने पर मिलती है। यदि $\angle\text{ATC} = 30^\circ$ और $\angle\text{ACT} = 48^\circ$ हो, तो $\angle\text{AOB}$ का मान क्या है? $\text{A}$ , $\text{B}$ and $\text{C}$ are three points on a circle with center $\text{O}$ . Tangent at $\text{C}$ meets $\text{BA}$ produced to $\text{T}$ . If $\angle\text{ATC} = 30^\circ$ and $\angle\text{ACT} = 48^\circ$ , then what is the value of $\angle\text{AOB}$ ? (SSC CGL Tier - I 2016) (a) $78^\circ$ (b) $96^\circ$ (c) $120^\circ$ (d) $108^\circ$

Answer: $108^\circ$

Step-by-step solution

Step 1: Apply Alternate Segment Theorem

According to the Alternate Segment Theorem, the angle between the tangent $TC$ and chord $AC$ at the point of contact $C$ is equal to the angle subtended by the chord $AC$ in the alternate segment. Therefore, $\angle\text{ABC} = \angle\text{ACT} = 48^\circ$ .

Step 2: Find angle CAT using triangle ATC

In $\triangle ATC$ , the sum of all interior angles is $180^\circ$ . Given that $\angle\text{ATC} = 30^\circ$ and $\angle\text{ACT} = 48^\circ$ , we find $\angle\text{CAT} = 180^\circ - 78^\circ = 102^\circ$ .

Step 3: Find interior angle CAB of triangle ABC

Points $T$ , $A$ , and $B$ lie on a straight line because side $BA$ is produced to $T$ . Hence, $\angle\text{BAC}$ and $\angle\text{CAT}$ form a linear pair, giving $\angle\text{BAC} = 180^\circ - 102^\circ = 78^\circ$ . Alternatively, $\angle\text{CAT}$ is an exterior angle to $\triangle ABC$ , so $\angle\text{CAT} = \angle\text{B} + \angle\text{C} = 102^\circ$ .

Step 4: Calculate angle ACB in triangle ABC

In $\triangle ABC$ , the sum of angles is $180^\circ$ . Using the values $\angle\text{BAC} = 78^\circ$ and $\angle\text{ABC} = 48^\circ$ , we calculate $\angle\text{ACB} = 180^\circ - 126^\circ = 54^\circ$ .

Step 5: Calculate angle AOB at the center

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Therefore, $\angle\text{AOB} = 2 \times \angle\text{ACB} = 2 \times 54^\circ = 108^\circ$ .

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