Question : AB is a chord in the minor segment of a circle with centre O. C is a point on the minor arc (between A and B ). The tangents to the circle at A and B meet at a point P. If $\angle \mathrm{ACB}=108^{\circ}$, then $\angle \mathrm{APB}$ is equal to:
Option 1: $36^{\circ}$
Option 2: $54^{\circ}$
Option 3: $27^{\circ}$
Option 4: $18^{\circ}$
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Correct Answer: $36^{\circ}$
Solution :
Point D is taken on the major arc of the circle. Then, AD, BD, CB, and AC are joined.$\angle ACB = 108^\circ$
Now,
$\angle ADB = (180 - 108)^\circ = 72^\circ$
Now,
$\angle AOB = (72\times 2)^\circ = 144^\circ$
Since $OA = OB,\triangle AOB$ is an isosceles triangle and $\angle OBA = \angle OAB$.
Now,
$\angle AOB + \angle OBA + \angle OAB = 180^\circ$
⇒ $2 \times \angle OBA = (180 - 144)^\circ$
⇒ $\angle OBA = 18^\circ$
Now,
$\angle OBA = \angle OAB = 18^\circ$
Now,
$\angle OAP = \angle OAB + \angle BAP$
⇒ $90^\circ = 18^\circ + \angle BAP$
⇒ $\angle BAP = 72^\circ$
Now,
$\angle OBP = \angle OBA + \angle ABP$
⇒ $90^\circ = 18^\circ + \angle ABP$
⇒ $\angle ABP = 72^\circ$
Now,
$\angle APB + \angle ABP + \angle BAP = 180^\circ$
⇒ $\angle APB + 72^\circ + 72^\circ = 180^\circ$
⇒ $\angle APB = 36^\circ$
Hence, the correct answer is $36^\circ$.
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