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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 Calculate Question number H_10 Adapted from N/A

Question

In triangle PQR, PR = 12  cm, QR = p  cm, PQ = r  cm and Q P ^ R = 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 r 2 12 3 r + 144 p 2 = 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.

p 2 = 12 2 + r 2 2 × 12 × r × cos ( 30 )      M1A1

r 2 12 3 r + 144 p 2 = 0      AG

[2 marks]

a.

EITHER

r 2 12 3 r + 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 = 1 2 × 12 × 5.1008 × sin ( 30 )      M1A1

= 15.3   ( c m 2 )      A1

[3 marks]

c.

METHOD 1

EITHER

r 2 12 3 r + 144 p 2 = 0

discriminant = ( 12 3 ) 2 4 × ( 144 p 2 )      M1

= 4 ( p 2 36 )      A1

( p 2 36 ) > 0      M1

p > 6      A1

OR

construction of a right angle triangle     (M1)

12 sin 30 = 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

6 < p < 12      A1

METHOD 2

diagram showing two triangles     (M1)

12 sin 30 = 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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