# IB DP Math AA Topic SL 3.2 :The cosine rule- IB Style Questions HL Paper 2

### Question: [Maximum mark: 6]

The following diagram shows a circle with centre O and radius 5 metres.
Points A and B lie on the circle and $$A\hat{O}B$$ = 1.9  radians.

(a) Find the length of the chord [AB].
(b) Find the area of the shaded sector.

Ans:

(a) EITHER
uses the cosine rule
AB2 = 52 + 52 -2 × 5 × 5 × cos1.9

OR
uses right-angled trigonometry

$$\frac{\frac{AB}{2}}{5}= sin 0.95$$

OR

uses the sine rule

$$\alpha \frac{1}{2}(\pi -1.9)(=0.6207…)$$

$$\frac{AB}{sin1.9}= \frac{5}{sin0.6207….}$$

THEN
AB = 8.1341…
AB = 8.13 (m)

(b) let the shaded area be A
METHOD 1
Attempt at finding reflex angle

$$A\hat{O}B = 2\pi -1.9 (=4.3831….)$$
substitution into area formula

$$A = \frac{1}{2}\times 5^{2}\times 4.3831…. OR \left ( \frac{2\pi -1.9}{2\pi } \right )\times \pi \left ( 5^{2} \right )$$
= 54.7898…
= 54.8 (m2)

METHOD 2
let the area of the circle be AC and the area of the unshaded sector be AU

## Question

In triangle ABC, BC = a , AC = b , AB = c and [BD] is perpendicular to [AC].

(a)     Show that $${\text{CD}} = b – c\cos A$$.

(b)     Hence, by using Pythagoras’ Theorem in the triangle BCD, prove the cosine rule for the triangle ABC.

(c)     If $${\rm{A\hat BC}} = 60^\circ$$ , use the cosine rule to show that $$c = \frac{1}{2}a \pm \sqrt {{b^2} – \frac{3}{4}{a^2}}$$ .[12]

Part A.

The above three dimensional diagram shows the points P and Q which are respectively west and south-west of the base R of a vertical flagpole RS on horizontal ground. The angles of elevation of the top S of the flagpole from P and Q are respectively 25° and 40° , and PQ = 20 m .

Determine the height of the flagpole.[8]

Part B.

## Markscheme

(a)     $${\text{CD}} = {\text{AC}} – {\text{AD}} = b – c\cos A$$     R1AG

[1 mark]

(b)     METHOD 1

$${\text{B}}{{\text{C}}^2} = {\text{B}}{{\text{D}}^2} + {\text{C}}{{\text{D}}^2}$$     (M1)

$${a^2} = {(c\sin A)^2} + {(b – c\cos A)^2}$$     (A1)

$$= {c^2}{\sin ^2}A + {b^2} – 2bc\cos A + {c^2}{\cos ^2}A$$     A1

$$= {b^2} + {c^2} – 2bc\cos A$$     A1

[4 marks]

METHOD 2

$${\text{B}}{{\text{D}}^2} = {\text{A}}{{\text{B}}^2} – {\text{A}}{{\text{D}}^2} = {\text{B}}{{\text{C}}^2} – {\text{C}}{{\text{D}}^2}$$     (M1)(A1)

$$\Rightarrow {c^2} – {c^2}{\cos ^2}A = {a^2} – {b^2} + 2bc\cos A – {c^2}{\cos ^2}A$$     A1

$$\Rightarrow {a^2} = {b^2} + {c^2} – 2bc\cos A$$     A1

[4 marks]

(c)     METHOD 1

$${b^2} = {a^2} + {c^2} – 2ac\cos 60^\circ \Rightarrow {b^2} = {a^2} + {c^2} – ac$$     (M1)A1

$$\Rightarrow {c^2} – ac + {a^2} – {b^2} = 0$$     M1

$$\Rightarrow c = \frac{{a \pm \sqrt {{{( – a)}^2} – 4({a^2} – {b^2})} }}{2}$$     (M1)A1

$$= \frac{{a \pm \sqrt {4{b^2} – 3{a^2}} }}{2} = \frac{a}{2} \pm \sqrt {\frac{{4{b^2} – 3{a^2}}}{4}}$$     (M1)A1

$$= \frac{1}{2}a \pm \sqrt {{b^2} – \frac{3}{4}{a^2}}$$     AG

Note: Candidates can only obtain a maximum of the first three marks if they verify that the answer given in the question satisfies the equation.

[7 marks]

METHOD 2

$${b^2} = {a^2} + {c^2} – 2ac\cos 60^\circ \Rightarrow {b^2} = {a^2} + {c^2} – ac$$     (M1)A1

$${c^2} – ac = {b^2} – {a^2}$$     (M1)

$${c^2} – ac + {\left( {\frac{a}{2}} \right)^2} = {b^2} – {a^2} + {\left( {\frac{a}{2}} \right)^2}$$     M1A1

$${\left( {c – \frac{a}{2}} \right)^2} = {b^2} – \frac{3}{4}{a^2}$$     (A1)

$$c – \frac{a}{2} = \pm \sqrt {{b^2} – \frac{3}{4}{a^2}}$$     A1

$$\Rightarrow c = \frac{1}{2}a \pm \sqrt {{b^2} – \frac{3}{4}{a^2}}$$     AG

[7 marks]

Part A.

$${\text{PR}} = h\tan 55^\circ {\text{ , QR}} = h\tan 50^\circ {\text{ where RS}} = h$$     M1A1A1

Use the cosine rule in triangle PQR.     (M1)

$${20^2} = {h^2}{\tan ^2}55^\circ + {h^2}{\tan ^2}50^\circ – 2h\tan 55^\circ h\tan 50^\circ \cos 45^\circ$$     A1

$${h^2} = \frac{{400}}{{{{\tan }^2}55^\circ + {{\tan }^2}50^\circ – 2\tan 55^\circ \tan 50^\circ \cos 45^\circ }}$$     (A1)

$$= 379.9…$$     (A1)

$$h = 19.5{\text{ (m)}}$$     A1

[8 marks]

## Question

In the right circular cone below, O is the centre of the base which has radius 6 cm. The points B and C are on the circumference of the base of the cone. The height AO of the cone is 8 cm and the angle $${\rm{B\hat OC}}$$ is 60°.

Calculate the size of the angle $${\rm{B\hat AC}}$$.

## Markscheme

AC = AB = 10 (cm)     A1

triangle OBC is equilateral     (M1)

BC = 6 (cm)     A1

EITHER

$${\rm{B\hat AC}} = 2\arcsin \frac{3}{{10}}$$     M1A1

$${\rm{B\hat AC}} = 34.9^\circ \,\,\,\,\,$$(accept 0.609 radians)     A1

OR

$$\cos {\rm{B\hat AC = }}\frac{{{{10}^2} + {{10}^2} – {6^2}}}{{2 \times 10 \times 10}} = \frac{{164}}{{200}}$$     M1A1

$${\rm{B\hat AC}} = 34.9^\circ \,\,\,\,\,$$(accept 0.609 radians)     A1

Note: Other valid methods may be seen.

[6 marks]

## Question

Consider the triangle ABC where $${\rm{B\hat AC}} = 70^\circ$$, AB = 8 cm and AC = 7 cm. The point D on the side BC is such that $$\frac{{{\text{BD}}}}{{{\text{DC}}}} = 2$$.

## Markscheme

use of cosine rule: $${\text{BC}} = \sqrt {({8^2} + {7^2} – 2 \times 7 \times 8\cos 70)} = 8.6426 \ldots$$     (M1)A1

Note: Accept an expression for $${\text{B}}{{\text{C}}^2}$$.

$${\text{BD}} = 5.7617 \ldots \,\,\,\,\,{\text{(CD}} = 2.88085 \ldots )$$     A1

use of sine rule: $$\hat B = \arcsin \left( {\frac{{7\sin 70}}{{{\text{BC}}}}} \right) = 49.561 \ldots ^\circ \,\,\,\,\,(\hat C = 60.4387 \ldots ^\circ )$$     (M1)A1

use of cosine rule: $${\text{AD}} = \sqrt {{8^2} + {\text{B}}{{\text{D}}^2} – 2 \times {\text{BD}} \times 8\cos B} = 6.12{\text{ (cm)}}$$     A1

Note: Scale drawing method not acceptable.

[6 marks]

## Question

Given $$\Delta$$ABC, with lengths shown in the diagram below, find the length of the line segment [CD].

## Markscheme

METHOD 1

$$\frac{{\sin C}}{7} = \frac{{\sin 40}}{5}$$     M1(A1)

$${\text{B}}\hat {\text{C}}{\text{D}} = 64.14…^\circ$$     A1

$${\text{CD}} = 2 \times 5\cos 64.14…$$     M1

Note: Also allow use of sine or cosine rule.

$${\text{CD}} = 4.36$$     A1

[5 marks]

METHOD 2

let $${\text{AC}} = x$$

cosine rule

$${5^2} = {7^2} + {x^2} – 2 \times 7 \times x\cos 40$$     M1A1

$${x^2} – 10.7 … x + 24 = 0$$

$$x = \frac{{10.7… \pm \sqrt {{{\left( {10.7…} \right)}^2} – 4 \times 24} }}{2}$$     (M1)

$$x = 7.54$$; $$3.18$$     (A1)

CD is the difference in these two values $$= 4.36$$     A1

Note: Other methods may be seen.

[5 marks

## Question

An electricity station is on the edge of a straight coastline. A lighthouse is located in the sea 200 m from the electricity station. The angle between the coastline and the line joining the lighthouse with the electricity station is 60°. A cable needs to be laid connecting the lighthouse to the electricity station. It is decided to lay the cable in a straight line to the coast and then along the coast to the electricity station. The length of cable laid along the coastline is x metres. This information is illustrated in the diagram below.

The cost of laying the cable along the sea bed is US$80 per metre, and the cost of laying it on land is US$20 per metre.

a.Find, in terms of x, an expression for the cost of laying the cable.[4]

b.Find the value of x, to the nearest metre, such that this cost is minimized.[2]

## Markscheme

let the distance the cable is laid along the seabed be y

$${y^2} = {x^2} + {200^2} – 2 \times x \times 200\cos 60^\circ$$     (M1)

(or equivalent method)

$${y^2} = {x^2} – 200x + 40000$$     (A1)

cost = C = 80y + 20x     (M1)

$$C = 80{({x^2} – 200x + 40000)^{\frac{1}{2}}} + 20x$$     A1

[4 marks]

a.

$$x = 55.2786 \ldots = 55$$ (m to the nearest metre)     (A1)A1

$$\left( {x = 100 – \sqrt {2000} } \right)$$

[2 marks]

Scroll to Top