Question
A triangle has sides of length \(({n^2} + n + 1)\), \((2n + 1)\) and \(({n^2} – 1)\) where \(n > 1\).
(a) Explain why the side \(({n^2} + n + 1)\) must be the longest side of the triangle.
(b) Show that the largest angle, \(\theta \), of the triangle is \(120^\circ \).
▶️Answer/Explanation
Markscheme
(a) a reasonable attempt to show either that \({n^2} + n + 1 > 2n + 1\) or
\({n^2} + n + 1 > {n^2} – 1\) M1
complete solution to each inequality A1A1
(b) \(\cos \theta = \frac{{{{(2n + 1)}^2} + {{({n^2} – 1)}^2} – {{({n^2} + n + 1)}^2}}}{{2(2n + 1)({n^2} – 1)}}\) M1A1
\( = \frac{{ – 2{n^3} – {n^2} + 2n + 1}}{{2(2n + 1)({n^2} – 1)}}\) M1
\( = – \frac{{(n – 1)(n + 1)(2n + 1)}}{{2(2n + 1)({n^2} – 1)}}\) A1
\( = – \frac{1}{2}\) A1
\(\theta = 120^\circ \) AG
[8 marks]
Question
In the triangle PQR, PQ = 6 , PR = k and \({\rm{P\hat QR}} = 30^\circ \) .
For the case k = 4 , find the two possible values of QR.[4]
Determine the values of k for which the conditions above define a unique triangle.[3]
▶️Answer/Explanation
Markscheme
attempt to apply cosine rule M1
\({4^2} = {6^2} + {\text{Q}}{{\text{R}}^2} – 2 \cdot {\text{QR}} \cdot 6\cos 30^\circ\) ( or \({\text{Q}}{{\text{R}}^2} – 6\sqrt 3 {\text{ QR}} + 20 = 0\) ) A1
\({\text{QR}} = 3\sqrt 3 + \sqrt 7 {\text{ or QR}} = 3\sqrt 3 – \sqrt 7 \) A1A1
[4 marks]
METHOD 1
\(k \geqslant 6\) A1
\(k = 6\sin 30^\circ = 3\) M1A1
Note: The M1 in (b) is for recognizing the right-angled triangle case.
METHOD 2
\(k \geqslant 6\) A1
use of discriminant: \(108 – 4(36 – {k^2}) = 0\) M1
k = 3 A1
Note: k = ±3 is M1A0.
[3 marks]
Question
The triangle ABC is equilateral of side 3 cm. The point D lies on [BC] such that BD = 1 cm.
Find \(\cos {\rm{D\hat AC}}\).
▶️Answer/Explanation
Markscheme
METHOD 1
\({\text{A}}{{\text{D}}^2} = {2^2} + {3^2} – 2 \times 2 \times 3 \times \cos 60^\circ \) M1
(or \({\text{A}}{{\text{D}}^2} = {1^2} + {3^2} – 2 \times 1 \times 3 \times \cos 60^\circ \))
Note: M1 for use of cosine rule with 60° angle.
\({\text{A}}{{\text{D}}^2} = 7\) A1
\(\cos {\rm{D\hat AC}} = \frac{{9 + 7 – 4}}{{2 \times 3 \times \sqrt 7 }}\) M1A1
Note: M1 for use of cosine rule involving \({\rm{D\hat AC}}\).
\( = \frac{2}{{\sqrt 7 }}\) A1
METHOD 2
let point E be the foot of the perpendicular from D to AC
EC = 1 (by similar triangles, or triangle properties) M1A1
(or AE = 2)
\({\text{DE}} = \sqrt 3 \) and \({\text{AD}} = \sqrt 7 \) (by Pythagoras) (M1)A1
\(\cos {\rm{D\hat AC}} = \frac{2}{{\sqrt 7 }}\) A1
Note: If first M1 not awarded but remainder of the question is correct award M0A0M1A1A1.
[5 marks]
Question
[without GDC]
In triangle ABC, AB = 9 cm, AC = 12 cm, and B is twice the size of C. Find the cosine of C.
▶️Answer/Explanation
Ans
\(\frac{9}{sinC}=\frac{12}{sin2C}\)
Using double angle formula \(\frac{9}{sinC}=\frac{12}{2sinCcosC}\)
\(\Rightarrow 9(2sinCcosC)=12sinC\)
\(\Rightarrow 6sinC(3cosC-2)=0\) or equivalent
(sin C ≠ 0)
\(\Rightarrow cos C=\frac{2}{3}\)
Question
[without GDC]
In triangle ABC, AB = 9 cm, AC = 12 cm, and B is twice the size of C. Find the cosine of C.
▶️Answer/Explanation
Ans
\(\frac{9}{sinC}=\frac{12}{sin2C}\)
Using double angle formula \(\frac{9}{sinC}=\frac{12}{2sinCcosC}\)
\(\Rightarrow 9(2sinCcosC)=12sinC\)
\(\Rightarrow 6sinC(3cosC-2)=0\) or equivalent
(sin C ≠ 0)
\(\Rightarrow cos C=\frac{2}{3}\)