A-level · Trigonometry

Arc length and sector areaExam-style questions

The radian formulas for arc length, sector area and segment area, where they come from, and using them in reverse — finding a radius or angle from a given perimeter or area, and the perimeters of sectors and segments.

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33 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4 Level 5

1
Figure 1

Diagram NOT accurately drawn

The design for a pendant is shown in Figure 1. It is made up of the triangle $ABC$ $A B C$ and the sector $ACD$ $A C D$ of a circle with centre $A$ $A$ , which share the edge $AC$ $A C$ .

Given that
- $AC = 5.6$ cm and $BC = 3.5$ cm
- angle $BAC = 0.5$ radians
- angle $ABC$ is obtuse
Making your method clear,
(a) calculate angle $ABC$ $A B C$ , giving your answer in radians to 4 significant figures,

(3 marks)
(b) calculate the area of triangle $ABC$ $A B C$ , giving your answer in cm $^2$ $^{2}$ to 3 significant figures.

(2 marks)

The total area of the pendant is $12$ $12$ cm $^2$ $^{2}$ .
(c) Show that angle $CAD = 0.537$ $C A D = 0.537$ radians, correct to 3 significant figures.

(3 marks)
(d) Hence find the perimeter of the pendant. Give your answer in centimetres, to the nearest whole number.

(5 marks)
Level 5 · 13 marksTry it
2
Figure 1

Diagram NOT accurately drawn

The plan view of a gravel path, $PQRS$ $P QR S$ , is shown in Figure 1.

Given that
- $OQR$ is a sector of a circle with centre $O$
- $OPS$ is a sector of a smaller circle, also with centre $O$ , where $P$ lies on $OQ$ and $S$ lies on $OR$
- angle $POS = 1.1$ radians
- the area of sector $OPS$ is $4.95$ m $^2$
(a) show that $OP = 3$ $O P = 3$ m

(2 marks)

The path is $x$ $x$ metres wide, so that $PQ = x$ $P Q = x$ m, and it covers an area of $40$ $40$ m $^2$ $^{2}$ .
(b) Show that

$11x^2 + 66x - 800 = 0$

(3 marks)
(c) Solve the equation in part (b) to find the value of $x$ $x$ , giving your answer to 3 significant figures.

(Solutions relying entirely on calculator technology are not acceptable.)

(2 marks)
(d) Hence find, to the nearest metre, the perimeter of the path.

(3 marks)
Level 5 · 10 marksTry it
3
The diagram shows a circular sector $OPQ$ , centre $O$ , with radius $r$ and angle $POQ = \theta$ radians, where $0 < \theta < \dfrac{\pi}{2}$

$N$ is the foot of the perpendicular from $P$ to $OQ$ (a) The area of triangle $OPN$ is $\dfrac{\sqrt{3}}{2}$ of the area of the sector $OPQ$

Show that

$\sin 2\theta = \sqrt{3}\,\theta$

(4 marks)
(b) Use a change of sign to show that the equation $\sin 2\theta = \sqrt{3}\,\theta$ has a root between $\dfrac{\pi}{8}$ and $\dfrac{\pi}{6}$

(2 marks)
(c) (i) Taking $\theta_1 = \dfrac{\pi}{6}$ as a first approximation, use the Newton-Raphson method to find $\theta_2$ and $\theta_3$

Give $\theta_3$ to three decimal places.

(ii) Explain how the Newton-Raphson method can be used to obtain $\theta$ correct to five decimal places.

(iii) A student uses $\theta_1 = \dfrac{\pi}{12}$ as the first approximation instead.

Explain why the method fails for this first approximation.

(6 marks)
Level 5 · 12 marksTry it
4
A flat metal weather vane is to be made in the shape shown in the diagram below.

The point $O$ lies on the straight line $PB$ , with $PO = 2OB$ .

The shape consists of the sector $OBD$ of a circle with centre $O$ and radius $OB$ , together with the triangle $OPD$ .

The angle $POD$ is $\theta$ radians, where $0 < \theta < \pi$ The length of $PB$ is $1.5$ metres.
(a) The area of the weather vane is $A$ square metres.

Show that

$A = \frac{1}{8}\left(\pi - \theta + 2\sin\theta\right)$

(4 marks)
(b) (i) Show that the area of the weather vane is greatest when $\theta = \dfrac{\pi}{3}$

Fully justify your answer.

(ii) Find, in exact form, the greatest possible area of the weather vane.

(8 marks)
(c) A second weather vane of the same design is made with $PB = 3$ metres.

Without further calculation, explain how this changes your answers to part (b).

(2 marks)
Level 5 · 14 marksTry it
5
Figure 1 — Diagram NOT accurately drawn

Figure 1 shows the plan of a lawn.

The lawn is made up of the sector $ORS$ $O R S$ of a circle with centre $O$ $O$ and the two triangles $OPQ$ $O P Q$ and $OVT$ $O V T$ . The triangles are congruent, and the points $R$ $R$ and $S$ $S$ lie on $OQ$ $O Q$ and $OT$ $O T$ respectively.

It is given that
- angle $ROS = 1.75$ radians
- the arc $RS$ has length $17.5$ m
- $P$ , $O$ and $V$ lie on a straight line, with $PV = 14$ m
(a) Show that the length of $OR$ $O R$ is $10$ $10$ m.

(2 marks)
(b) Show that angle $POQ$ $P O Q$ is $0.696$ $0.696$ radians, correct to 3 decimal places.

(2 marks)

The boundary of the lawn from $Q$ $Q$ to $T$ $T$ is made up of the straight edge $QR$ $QR$ , the arc $RS$ $R S$ and the straight edge $ST$ $S T$ . This part of the boundary has a total length of $23.5$ $23.5$ m.
(c) Find the area of the lawn, giving your answer to the nearest square metre.

(6 marks)
Level 5 · 10 marksTry it

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