# IBDP Maths analysis and approaches Topic: SL 3.2 : The sine rule including the ambiguous case HL Paper 1

### Question

In triangle ABC, AB = 9 cm , AC = 12 cm , and $$\hat B$$ is twice the size of $${\hat C}$$ .

Find the cosine of $${\hat C}$$ .

### Markscheme

$$\frac{9}{{\sin C}} = \frac{{12}}{{\sin B}}$$     (M1)

$$\frac{9}{{\sin C}} = \frac{{12}}{{\sin 2C}}$$     A1

Using double angle formula $$\frac{9}{{\sin C}} = \frac{{12}}{{2\sin C\cos C}}$$     M1

$$\Rightarrow 9(2\sin C\cos C) = 12\sin C$$

$$\Rightarrow 6\sin C(3\cos C – 2) = 0\,\,\,\,\,{\text{or equivalent}}$$     (A1)

$$(\sin C \ne 0)$$

$$\Rightarrow \cos C = \frac{2}{3}$$     A1

[5 marks]

### Question

In the triangle PQR, PQ = 6 , PR = k and $${\rm{P\hat QR}} = 30^\circ$$ .

a.For the case k = 4 , find the two possible values of QR.[4]

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

### 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

In triangle $${\text{ABC, BC}} = \sqrt 3 {\text{ cm}}$$, $${\rm{A\hat BC}} = \theta$$ and $${\rm{B\hat CA}} = \frac{\pi }{3}$$.

a.Show that length $${\text{AB}} = \frac{3}{{\sqrt 3 \cos \theta + \sin \theta }}$$.[4]

b.Given that $$AB$$ has a minimum value, determine the value of $$\theta$$ for which this occurs.[4]

## Markscheme

any attempt to use sine rule     M1

$$\frac{{{\text{AB}}}}{{\sin \frac{\pi }{3}}} = \frac{{\sqrt 3 }}{{\sin \left( {\frac{{2\pi }}{3} – \theta } \right)}}$$     A1

$$= \frac{{\sqrt 3 }}{{\sin \frac{{2\pi }}{3}\cos \theta – \cos \frac{{2\pi }}{3}\sin \theta }}$$     A1

Note:     Condone use of degrees.

$$= \frac{{\sqrt 3 }}{{\frac{{\sqrt 3 }}{2}\cos \theta + \frac{1}{2}\sin \theta }}$$     A1

$$\frac{{{\text{AB}}}}{{\frac{{\sqrt 3 }}{2}}} = \frac{{\sqrt 3 }}{{\frac{{\sqrt 3 }}{2}\cos \theta + \frac{1}{2}\sin \theta }}$$

$$\therefore {\text{AB}} = \frac{3}{{\sqrt 3 \cos \theta + \sin \theta }}$$     AG

[4 marks]

a.

METHOD 1

$$({\text{AB}})’ = \frac{{ – 3\left( { – \sqrt 3 \sin \theta + \cos \theta } \right)}}{{{{\left( {\sqrt 3 \cos \theta + \sin \theta } \right)}^2}}}$$     M1A1

setting $$({\text{AB}})’ = 0$$     M1

$$\tan \theta = \frac{1}{{\sqrt 3 }}$$

$$\theta = \frac{\pi }{6}$$     A1

METHOD 2

$${\text{AB}} = \frac{{\sqrt 3 \sin \frac{\pi }{3}}}{{\sin \left( {\frac{{2\pi }}{3} – \theta } \right)}}$$

$$AB$$ minimum when $$\sin \left( {\frac{{2\pi }}{3} – \theta } \right)$$ is maximum     M1

$$\sin \left( {\frac{{2\pi }}{3} – \theta } \right) = 1$$     (A1)

$$\frac{{2\pi }}{3} – \theta = \frac{\pi }{2}$$     M1

$$\theta = \frac{\pi }{6}$$     A1

METHOD 3

shortest distance from $$B$$ to $$AC$$ is perpendicular to $$AC$$     R1

$$\theta = \frac{\pi }{2} – \frac{\pi }{3} = \frac{\pi }{6}$$     M1A2

[4 marks]

Total [8 marks]

b.

### Question

The following diagram shows the triangle ABC where $${\text{AB}} = 2,{\text{ AC}} = \sqrt 2$$ and $${\rm{B\hat AC}} = 15^\circ$$.

a.Expand and simplify $${\left( {1 – \sqrt 3 } \right)^2}$$.[1]

b.By writing $$15^\circ$$ as $$60^\circ – 45^\circ$$ find the value of $$\cos (15^\circ )$$.[3]

c.Find BC in the form $$a + \sqrt b$$ where $$a,{\text{ }}b \in \mathbb{Z}$$.[4]

### Markscheme

$${\left( {1 – \sqrt 3 } \right)^2} = 4 – 2\sqrt 3$$    A1

Note:     Award A0 for $$1 – 2\sqrt 3 + 3$$.

[1 mark]

a.

$$\cos (60^\circ – 45^\circ ) = \cos (60^\circ )\cos (45^\circ ) + \sin (60^\circ )\sin (45^\circ )$$    M1

$$= \frac{1}{2} \times \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 3 }}{2} \times \frac{{\sqrt 2 }}{2}{\text{ }}\left( {{\text{or }}\frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}} \right)$$    (A1)

$$= \frac{{\sqrt 2 + \sqrt 6 }}{4}{\text{ }}\left( {{\text{or }}\frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}} \right)$$    A1

[3 marks]

b.

$$B{C^2} = 2 + 4 – 2 \times \sqrt 2 \times 2\cos (15^\circ )$$    M1

$$= 6 – \sqrt 2 \left( {\sqrt 2 + \sqrt 6 } \right)$$

$$= 4 – \sqrt {12} {\text{ }}\left( { = 4 – 2\sqrt 3 } \right)$$    A1

$$BC = \pm \left( {1 – \sqrt 3 } \right)$$    (M1)

$$BC = – 1 + \sqrt 3$$    A1

Note:     Accept $$BC = \sqrt 3 – 1$$.

Note:     Award M1A0 for $$1 – \sqrt 3$$.

Note:     Valid geometrical methods may be seen.

[4 marks]

c.
Scroll to Top