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TMUA 2017 · Paper 1 · Question 20 of 20

TMUA 2017 Paper 1 Question 20

Trigonometry — Cosine rule · range of an angle under a constraint. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 1TrigonometryCosine rule · range of an angle under a constraint5 optionshard
The lengths of the sides QR, RP and PQ in triangle PQR are a, a + d and a + 2d respectively, where a and d are positive and such that 3d > 2a.

What is the full range, in degrees, of possible values for angle PRQ?

  1. A0 < angle  PRQ < 60
  2. B0 < angle  PRQ < 120
  3. C60 < angle  PRQ < 120
  4. D60 < angle  PRQ < 180
  5. E120 < angle  PRQ < 180
Show the answer and worked solution
answer · E
  1. A0 < angle  PRQ < 60
  2. B0 < angle  PRQ < 120
  3. C60 < angle  PRQ < 120
  4. D60 < angle  PRQ < 180
  5. E120 < angle  PRQ < 180
Angle PRQ is at R, opposite the side PQ = a + 2d, so the cosine rule gives cos R = a2 + (a+d)2 − (a+2d)22a(a+d). The numerator simplifies to a2 − 2ad − 3d2 = (a − 3d)(a + d), so cos R = a − 3d2a. There are two constraints on d: the triangle inequality a + (a+d) > a + 2d forces d < a, and the given condition 3d > 2a forces d > 2a3. Writing t = da, so 23 < t < 1, gives cos R = 1 − 3t2, which runs strictly between −1 and −12. Hence 120 < R < 180.