Question
Download Solution PDFFrom an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB= 55°, find ∠AOB.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
PA and PB are the tangents to the circle with centre O.
∠PAB= 55°
Concept:
Tangents drawn from the same external point are equal in length.
A tangent perpendicular to the radius at the point of tangency.
Calculation:
∵ ∠PAB = 55°
∴ ∠PBA = 55° (PA = PB)
In triangle PAB,
∠APB + ∠PAB + ∠PBA = 180° (Angle Sum Property)
⇒ ∠P + 55° + 55° = 180°
⇒ ∠P = 70°
Also, ∠AOB + ∠APB = 180° (Sum of all the angles of a quadrilateral is 360° & ∠P = 70° ∠B = 90°)
⇒ ∠AOB = 180° - 70° = 110
∴ The measure of ∠AOB = 110°
Last updated on Jul 5, 2025
-> RRB NTPC Under Graduate Exam Date 2025 has been released on the official website of the Railway Recruitment Board.
-> The RRB NTPC Admit Card will be released on its official website for RRB NTPC Under Graduate Exam 2025.
-> Candidates who will appear for the RRB NTPC Exam can check their RRB NTPC Time Table 2025 from here.
-> The RRB NTPC 2025 Notification released for a total of 11558 vacancies. A total of 3445 Vacancies have been announced for Undergraduate posts like Commercial Cum Ticket Clerk, Accounts Clerk Cum Typist, Junior Clerk cum Typist & Trains Clerk.
-> A total of 8114 vacancies are announced for Graduate-level posts in the Non-Technical Popular Categories (NTPC) such as Junior Clerk cum Typist, Accounts Clerk cum Typist, Station Master, etc.
-> Prepare for the exam using RRB NTPC Previous Year Papers.
-> Get detailed subject-wise UGC NET Exam Analysis 2025 and UGC NET Question Paper 2025 for shift 1 (25 June) here