Question

please solve urgeng​

Answer 1

✧ Q.30. :-

Since, O is the centre and r is the radius of the circle and PS and PT are the tangents to the circle.

✧ To proof :-

$\bf \angle \: OTS \: = \angle \: OST = 30 \degree$

$\bf \huge{ \mathfrak{ \underline{ \underline{ \purple P \green r \orange o \blue o \pink f :- }}}}$

As, PT and PS are tangents drawn in the circle.

$\bf \implies \angle \: OTP \: = \angle \: OSP\: = 90 \degree \: \\ \bf \: (tangents \: to \: the \: circle \: is \: \perp \: to \: the \: radius \: to \: the \: point \: of \: contact)$

$\rm\green{\underline{\underline{\bold{✧In \: \triangle \: OTP,}}}}$

$\bf \: sin \angle \: P</p><p>OPT \: = \frac{OT}{OP} = \frac{x}{2x} \\ \bf \therefore \: sin \angle \: OPT \: = \frac{1}{2} \\ \bf \: so \: \angle \: OPT \: = 3 0\degree \\ \bf \: then \: \angle \: OTP + \angle \:OPT+ \angle \: TOP= 180 \degree \\ \bf \: (by \: angle \: sum \: property) \\ \bf \: \implies \: 90 \degree+ 30 \degree \: + \angle \: top \: = 180 \degree \\ \bf \: \therefore \angle \: top \: = 180 \degree - 120 \degree =60 \degree \\ \:$

$\rm\blue{\underline{\underline{\bold{✧ Now,\:\:∆PTS}}}}$

Is an Isoceles ∆ and OP is the angle bisector of TPS.

$\bf \therefore \: TS \: \perp OP \\ \bf \: \implies \: \angle \: OQT \: \angle \: OQS\: = 90 \degree \\ \bf \: in \triangle OTQ\: \\ \bf \: \implies \: \angle \: OQT\: + \angle \: OQT\angle + \: TOQ \: = 180 \degree \: \\ \bf \implies \: 90 \degree + \angle \: OTQ + 60\degree = 180 \degree \: \\ \bf \therefore \angle \: OTQ \: = 180 \degree- 150 \degree = 30 \degree$

Similarly, :-

$\bf{\orange {\underline{\underline{\angle \: OSQ\: = \angle \: OST \: = 30 \degree\:\:☞\:ans.}}}}$

$\bf \huge{\rm{ \underline{ \underline{ Hence, \:\purple P \green r \orange o \blue o \pink v \blue e \purple d}}}}$