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TMUA 2018 · Paper 2 · Question 14 of 20

TMUA 2018 Paper 2 Question 14

Trigonometry — The ambiguous case · when SSA fixes the third side. Try it first; the answer and a full worked solution are below.

TMUA 2018 · Paper 2TrigonometryThe ambiguous case · when SSA fixes the third side8 optionshard
In the triangle PQR, PR = 2, QR = p and ∠ RPQ = 30∘.

What is the set of all the values of p for which this information uniquely determines the length of PQ?

  1. Ap = 1
  2. Bp = √3
  3. C1 ≤ p < 2
  4. D√3 ≤ p < 2
  5. Ep = 1 or p ≥ 2
  6. Fp = √3 or p ≥ 2
  7. Gp < 2
  8. Hp ≥ 2
Show the answer and worked solution
answer · E
  1. Ap = 1
  2. Bp = √3
  3. C1 ≤ p < 2
  4. D√3 ≤ p < 2
  5. Ep = 1 or p ≥ 2
  6. Fp = √3 or p ≥ 2
  7. Gp < 2
  8. Hp ≥ 2
This is the ambiguous SSA configuration, so work with the cosine rule and count roots. Writing c = PQ, the rule at P gives p2 = c2 + 4 − 4ccos 30∘, that is c2 − 2√3 c + (4 − p2) = 0, so c = √3 ± √p2 − 1. For a triangle we need a real, strictly positive c. If p < 1 there is none; if p = 1 the two roots coincide at c = √3, giving exactly one triangle. For 1 < p < 2 both roots are positive, so two triangles fit and PQ is not determined. At p = 2 the smaller root is 0 and for p > 2 it is negative, leaving one valid triangle. So the answer is p = 1 or p ≥ 2.