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Date May 2017 Marks available 3 Reference code 17M.2.AHL.TZ1.H_10
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Find and Hence Question number H_10 Adapted from N/A

Question

In triangle PQR, PR=12 cm, QR=p cm, PQ=r cm and QˆPR=30.

Consider the possible triangles with QR=8 cm.

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

Use the cosine rule to show that r2123r+144p2=0.

[2]
a.

Calculate the two corresponding values of PQ.

[3]
b.

Hence, find the area of the smaller triangle.

[3]
c.

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

[7]
d.

Markscheme

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

p2=122+r22×12×r×cos(30)     M1A1

r2123r+144p2=0     AG

[2 marks]

a.

EITHER

r2123r+80=0     (M1)

OR

using the sine rule     (M1)

THEN

PQ=5.10 (cm) or     A1

PQ=15.7 (cm)     A1

[3 marks]

b.

area=12×12×5.1008×sin(30)     M1A1

=15.3 (cm2)     A1

[3 marks]

c.

METHOD 1

EITHER

r2123r+144p2=0

discriminant =(123)24×(144p2)     M1

=4(p236)     A1

(p236)>0     M1

p>6     A1

OR

construction of a right angle triangle     (M1)

12sin30=6     M1(A1)

hence for two triangles p>6     R1

THEN

p<12     A1

144p2>0 to ensure two positive solutions or valid geometric argument     R1

6<p<12     A1

METHOD 2

diagram showing two triangles     (M1)

12sin30=6     M1A1

one right angled triangle when p=6     (A1)

p>6 for two triangles     R1

p<12 for two triangles     A1

6<p<12     A1

[7 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.2—2d and 3d trig
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Topic 3—Geometry and trigonometry » AHL 3.8—Unit circle, Pythag identity, solving trig equations graphically
Topic 3—Geometry and trigonometry

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