Exams › JEE Advanced › Maths
In an acute-angled triangle PQR, let O be the orthocentre and C the circumcentre. The line PO is extended to meet side QR at S. Given angle QCR = 130 deg and angle PQS = 60 deg, find angle RPS.
- 35 deg
- 30 deg
- 50 deg
- 60 deg
Correct answer: 35 deg
Solution
Central angle QCR = 2 * angle QPR => angle QPR = 130/2 = 65 deg. Since O (orthocentre) lies on the altitude from P, line PO meets QR perpendicularly, so angle PSQ = 90 deg. In triangle PSQ: angle QPS = 180 - 90 - 60 = 30 deg. Then angle RPS = angle QPR - angle QPS = 65 - 30 = 35 deg.
Related JEE Advanced Maths questions
- In a chemistry class, there are 20 students, while a physics class has 30 students. If 10 students are enrolled in both classes, and the two classes are held at separate times, determine the value of k/8, where k represents the total number of students attending either class.
- Given that A represents the divisors of 15, B contains prime numbers less than 10, and C includes even numbers less than 9, how many elements are there in the intersection of (A ∪ C) and B?
- Identify the periodic function among the following:
- The function f(x) = √|x² - 5| x + 6 + √8 + 2|x| - |x|² is defined as a real number for values of x within which range?
- Let f(x) = 4x / (4x² + 2). If the sum of the integrals ∫(1/1997) + ∫(2/1997) +... + ∫(1196/1997) equals 499q, what is the value of q?
- Given that α, β, γ, and θ are the smallest positive angles in increasing order for which their sine values equal a positive constant k, what is the result of 4 sin(α/2) + 3 sin(β/2) + 2 sin(γ/2) + sin(θ/2)?
⚔️ Practice JEE Advanced Maths free + battle 1v1 →