3.29: Triangles with exact trig ratios

0606/11/M/J/23 — Question 5 · 8 marks

Do not use a calculator in this question.

In this question, all lengths are in centimetres.
(a)
Given cos 120∘=−12\mathrm{cos}\,120^{\circ} = -\tfrac{1}{2}. In triangle ABCABC, AB=53−6AB = 5\sqrt{3} - 6, BC=53+6BC = 5\sqrt{3} + 6 and ∠ABC=120∘\angle ABC = 120^{\circ}. Find ACAC in the form aba\sqrt{b}.
[4]
(b)
Given sin 30∘=12\mathrm{sin}\,30^{\circ} = \tfrac{1}{2}. In triangle PQRPQR, PQ=3+25PQ = 3 + 2\sqrt{5} and ∠PQR=30∘\angle PQR = 30^{\circ}. The area is 5+254\dfrac{5 + 2\sqrt{5}}{4}. Find QRQR in the form c+d5c + d\sqrt{5}.
[4]