Ex.10.2 Q2 Circles Solution - NCERT Maths Class 10

Go back to  'Ex.10.2'

Question

In the given figure, if \(TP\) and \(TQ\) are the two tangents to a circle with centre \(O\) so that \(\angle {P O Q} = 110 ^ { \circ }\), then \(\angle {P T Q} \) is equal to

(A)\(60^\circ\) 

(B)\(70^\circ\)  

(C)\(80^\circ \) 

(D)\(90^\circ\)

 Video Solution
Circles
Ex 10.2 | Question 2

Text Solution

What is Known?

(i) \(TP\) and \(TQ\) are tangents to a circle with Centre \(O\)

(ii) \(\angle {P O Q}=110^{\circ}\)

What is Unknown?

\(\angle {PTQ}\)

Reasoning:

  • Tangent at any point of a circle is  perpendicular to the radius through the point of contact.
  • In the above figure \(OPTQ\) is a quadrilateral and \(\angle {P}\) and \(\angle {Q}\)  are \(90^{\circ}\) 
  • Sum of the angles of a quadrilateral is \(360^{\circ}\)

Steps:

\(\therefore \;\)In \(OPTQ,\)

\[\begin{align}   \left[ \begin{array}  & \angle Q+\angle P+ \\  \angle POQ+\angle PTQ \\ \end{array} \right]&\!=\!{{360}^{{}^\circ }} \\  \left[ \begin{array}  & {{90}^{{}^\circ }}+{{90}^{{}^\circ }}+ \\ 
 {{110}^{{}^\circ }}+\angle PTQ \\ \end{array} \right]&\!=\!{{360}^{{}^\circ }} \\   {{290}^{{}^\circ }}+\angle PTQ&\!=\!{{360}^{{}^\circ }} \\   \angle PTQ&\!=\!{{360}^{{}^\circ }}-{{290}^{{}^\circ }} \\ \angle PTQ&\!=\!{{70}^{{}^\circ }} \\ \end{align}\]

Hence the correct Option is B

  
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school