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 $$120^\circ$$.

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.

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 $$\cos {\rm{D\hat AC}}$$.

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.

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.

$$\frac{9}{sinC}=\frac{12}{sin2C}$$
Using double angle formula $$\frac{9}{sinC}=\frac{12}{2sinCcosC}$$
$$\Rightarrow 9(2sinCcosC)=12sinC$$
$$\Rightarrow 6sinC(3cosC-2)=0$$ or equivalent
$$\Rightarrow cos C=\frac{2}{3}$$