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 r2−12√3r+144−p2=0.
Calculate the two corresponding values of PQ.
Hence, find the area of the smaller triangle.
Determine the range of values of p for which it is possible to form two triangles.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
p2=122+r2−2×12×r×cos(30∘) M1A1
r2−12√3r+144−p2=0 AG
[2 marks]
EITHER
r2−12√3r+80=0 (M1)
OR
using the sine rule (M1)
THEN
PQ=5.10 (cm) or A1
PQ=15.7 (cm) A1
[3 marks]
area=12×12×5.1008…×sin(30∘) M1A1
=15.3 (cm2) A1
[3 marks]
METHOD 1
EITHER
r2−12√3r+144−p2=0
discriminant =(12√3)2−4×(144−p2) M1
=4(p2−36) A1
(p2−36)>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
144−p2>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]