Question : $\text{PS}$ and $\text{PT}$ are two tangents from a point $\text{P}$ outside the circle with centre $\text{O}$. If $\text{S}$ and $\text{T}$ are points on the circle such that $\angle SPT=130^{\circ}$, then the degree measure of $\angle OST$ is equal to:
Option 1: $25^{\circ}$
Option 2: $55^{\circ}$
Option 3: $65^{\circ}$
Option 4: $35^{\circ}$
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Correct Answer: $65^{\circ}$
Solution :
$\angle SPT = 130^{\circ}$
And also,
$\angle PSO = \angle PTO = 90^{\circ}$
Now, PTOS is a quadrilateral.
$\therefore$ $\angle P + \angle O + \angle S + \angle P = 360^{\circ}$
⇒ $\angle O = 360^{\circ} - 310^{\circ} = 50°$
Now to calculate $\angle OST$.
$\angle OST$ will be equal to $\angle OTS$.
In $\triangle OST$,
$\angle OST + \angle OTS + \angle TOS = 180^{\circ}$
⇒ $2\angle OST = 180^{\circ} - 50^{\circ}=130^{\circ}$
$\therefore\angle OST = 65^{\circ}$
Hence, the correct answer is $65^{\circ}$.
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