IBDP Maths analysis and approaches Topic: SL 3.2 :The cosine rule HL Paper 1

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]

a.

Determine the values of k for which the conditions above define a unique triangle.[3]

b.
▶️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]

a.

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]

b.

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)
\(\Rightarrow cos C=\frac{2}{3}\)

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