17M.2.AHL.TZ1.H_10

pestleMathematicsAIHLPaper 217M· sl-3-2-2d-and-3d-trigsource ↗

In triangle ${\text{PQR, PR}} = 12{\text{ cm, QR}} = p{\text{ cm, PQ}} = r{\text{ cm}}$ and ${\rm{Q\hat PR}} = 30^\circ$ .

Consider the possible triangles with ${\text{QR}} = 8{\text{ cm}}$ .

Consider the case where $p$ , the length of QR is not fixed at 8 cm.

Use the cosine rule to show that ${r^2} - 12\sqrt 3 r + 144 - {p^2} = 0$ .

[2]

Calculate the two corresponding values of PQ.

[3]

Hence, find the area of the smaller triangle.

[3]

Determine the range of values of $p$ for which it is possible to form two triangles.

[7]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${p^2} = {12^2} + {r^2} - 2 \times 12 \times r \times \cos (30^\circ )$ M1A1

${r^2} - 12\sqrt 3 r + 144 - {p^2} = 0$ AG

[2 marks]

EITHER

${r^2} - 12\sqrt 3 r + 80 = 0$ (M1)

using the sine rule (M1)

THEN

${\text{PQ}} = 5.10{\text{ }}({\text{cm}})$ or A1

${\text{PQ}} = 15.7{\text{ }}({\text{cm}})$ A1

[3 marks]

${\text{area}} = \frac{1}{2} \times 12 \times 5.1008 \ldots \times \sin (30^\circ )$ M1A1

$= 15.3{\text{ }}({\text{c}}{{\text{m}}^2})$ A1

[3 marks]

METHOD 1

EITHER

${r^2} - 12\sqrt 3 r + 144 - {p^2} = 0$

discriminant $= {\left( {12\sqrt 3 } \right)^2} - 4 \times (144 - {p^2})$ M1

$= 4({p^2} - 36)$ A1

$({p^2} - 36) > 0$ M1

$p > 6$ A1

construction of a right angle triangle (M1)

$12\sin 30^\circ = 6$ M1(A1)

hence for two triangles $p > 6$ R1

THEN

$p < 12$ A1

$144 - {p^2} > 0$ to ensure two positive solutions or valid geometric argument R1

$\therefore 6 < p < 12$ A1

METHOD 2

diagram showing two triangles (M1)

$12\sin 30^\circ = 6$ M1A1

one right angled triangle when $p = 6$ (A1)

$\therefore p > 6$ for two triangles R1

$p < 12$ for two triangles A1

$6 < p < 12$ A1

[7 marks]

Examiners’ report

[N/A]