What is 5.9: Trig in context?

Resolve a vector of magnitude V at angle θ to the positive horizontal into components Vcosθ and Vsinθ, changing signs to match its actual direction.

How many marks are 5.9 questions worth?

5.9 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 5.9 appeared in?

We have not yet indexed a verified past-paper appearance for 5.9 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

5.9

Use trigonometric functions to solve problems in context, including problems involving vectors, kinematics and forces.

Draft — not yet indexed

Trig in context

Worked answers and methods for 5.9 on Edexcel A-level Maths 9MA0.

Explanation

Resolve a vector of magnitude $V$ at angle $\theta$ to the positive horizontal into components $V\cos\theta$ and $V\sin\theta$ , changing signs to match its actual direction.
For resultant or equilibrium problems, form separate equations in two perpendicular directions; a zero resultant requires both component sums to equal zero.
Trigonometric models can describe wave motion, a point on a vertical circular wheel or changing hours of sunlight.
Interpret every solution using the stated time interval, units and physical constraints.
A calculator angle without a quadrant check can point in the opposite direction; draw and label a diagram, then state bearings or directions in the form the context requests.

Perpendicular components combine to give a resultant whose direction is measured from east.

Worked example

Two horizontal forces act on a crate: $8\,\text{N}$ due east and $11\,\text{N}$ at $60^\circ$ north of east. Find the magnitude and direction of their resultant, to $3$ significant figures and the nearest degree respectively.

1.The east component is $8+11\cos60^\circ=13.5\,\text{N}$ and the north component is $11\sin60^\circ=9.526\ldots\,\text{N}$ .
2.Hence $R=\sqrt{13.5^2+9.526^2}=16.523\ldots\,\text{N}$ .
3.Its direction is $\arctan(9.526/13.5)=35.21\ldots^\circ$ north of east.

Answer: Magnitude $=16.5\,\text{N}$ Direction $35^\circ$ north of east

Common mistakes

Don't use an inverse-tangent calculator value without checking the resultant's quadrant.
Don't state a bare angle without the bearing or directional wording required by the context.

Exam tip

Draw and label component directions, then give the final magnitude with units and the direction in contextual form.

Worked practice

Tier 1 · Easy

A drone travels at

12\,\text{m s}^{-1}

on a path

35^\circ

above the horizontal. Calculate its horizontal and vertical velocity components to

3

significant figures.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Horizontal $=9.83\,\text{m s}^{-1}$ Vertical $=6.88\,\text{m s}^{-1}$	3
Notes
Resolve the velocity vector: the horizontal component is $12\cos35^\circ=9.829\ldots\,\text{m s}^{-1}$ and the vertical component is $12\sin35^\circ=6.882\ldots\,\text{m s}^{-1}$ .

(3 marks)

Tier 2 · Standard

A hiker walks

24\,\text{km}

on a bearing of

040^\circ

and then

18\,\text{km}

on a bearing of

130^\circ

. Find the hiker's distance and bearing from the starting point, giving the bearing to the nearest degree.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Distance $=30\,\text{km}$ Bearing $077^\circ$	5
Notes
Resolve each displacement into east and north components. The totals are $E=24\sin40^\circ+18\sin130^\circ=29.2157\ldots$ and $N=24\cos40^\circ+18\cos130^\circ=6.81489\ldots$ . Thus the distance is $\sqrt{E^2+N^2}=30\,\text{km}$ . The bearing is measured clockwise from north, so it is $\arctan(E/N)=76.869\ldots^\circ$ , giving $077^\circ$ .

(5 marks)

Tier 3 · Hard

A ring is in equilibrium under its weight of

18\,\text{N}

, a tension

P

directed

25^\circ

above the horizontal, and a horizontal tension

Q

acting oppositely. Calculate

P

and

Q

3

significant figures.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$P=42.6\,\text{N}$ $Q=38.6\,\text{N}$	5
Notes
Vertical equilibrium gives $P\sin25^\circ=18$ , so $P=18/\sin25^\circ=42.592\ldots\,\text{N}$ . Horizontal equilibrium then gives $Q=P\cos25^\circ=38.601\ldots\,\text{N}$ . Therefore $P=42.6\,\text{N}$ and $Q=38.6\,\text{N}$ .

(5 marks)

Tier 1 · Easy

The height of a marker above the floor is modelled by

h=5+3\sin(\pi t/6)

metres, where

t

is measured in seconds. Find the exact height when

t=2

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$5+\dfrac{3\sqrt3}{2}\,\text{m}$	2
Notes
Substitute $t=2$ : $h=5+3\sin(\pi/3)=5+3(\sqrt3/2)=5+3\sqrt3/2\,\text{m}$ .

(2 marks)

Tier 2 · Standard

The position vector of a particle is

\mathbf r=(6\cos t)\mathbf i+(8\sin t)\mathbf j

metres. When

t=\pi/3

, find the particle's distance from the origin exactly and the direction of

\mathbf r

, in degrees above the positive

x

-axis, to

1

decimal place.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Distance $=\sqrt{57}\,\text{m}$ Direction $=66.6^\circ$	4
Notes
At $t=\pi/3$ , $\mathbf r=3\mathbf i+4\sqrt3\mathbf j$ . Its magnitude is $\sqrt{3^2+(4\sqrt3)^2}=\sqrt{57}\,\text{m}$ . Both components are positive, so the direction is $\arctan(4\sqrt3/3)=66.586\ldots^\circ$ , giving $66.6^\circ$ above the positive $x$ -axis.

(4 marks)

Tier 3 · Hard

An aircraft is required to have a ground velocity of

120\,\text{km h}^{-1}

due north. A steady wind has velocity

30\,\text{km h}^{-1}

due east. Find the constant airspeed and bearing that the pilot must use, giving the airspeed to

3

significant figures and the bearing to the nearest degree.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Airspeed $=124\,\text{km h}^{-1}$ Bearing $346^\circ$	5
Notes
Use $\mathbf v_{\text{air}}+\mathbf v_{\text{wind}}=\mathbf v_{\text{ground}}$ . Taking east and north as positive component directions, $\mathbf v_{\text{air}}=(-30,120)$ . Hence the airspeed is $\sqrt{30^2+120^2}=30\sqrt{17}=123.693\ldots\,\text{km h}^{-1}$ , giving $124\,\text{km h}^{-1}$ . The direction is $\arctan(30/120)=14.036\ldots^\circ$ west of north, so the bearing measured clockwise from north is $360^\circ-14.036\ldots^\circ=345.963\ldots^\circ$ , giving $346^\circ$ .

(5 marks)

Tier 2 · Standard

Points

A

and

B

are

10\,\text{km}

apart, with

B

due east of

A

. A buoy

C

is on a bearing of

040^\circ

from

A

and a bearing of

310^\circ

from

B

. Find the distances

AC

and

BC

, giving each answer to

3

significant figures.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$AC=6.43\,\text{km}$ $BC=7.66\,\text{km}$	4
Notes
The bearing of $B$ from $A$ is $090^\circ$ , so $\angle BAC=50^\circ$ . The bearing of $A$ from $B$ is $270^\circ$ , so $\angle ABC=40^\circ$ . Hence $\angle ACB=90^\circ$ . By the sine rule, $AC=10\sin40^\circ=6.427\ldots\,\text{km}$ and $BC=10\sin50^\circ=7.660\ldots\,\text{km}$ , giving the stated answers.

(4 marks)

Tier 3 · Hard

A sign of weight

W

newtons is held in equilibrium by two light cables. The left cable runs upwards to the left at

30^\circ

above the horizontal and can withstand a maximum tension of

80\,\text{N}

. The right cable runs upwards to the right at

60^\circ

above the horizontal and can withstand a maximum tension of

96\,\text{N}

. The sign is modelled as a particle. Find the greatest possible value of

W

, giving your answer to

3

significant figures, and state which cable is at its limiting tension. Use unrounded values in your working.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$W=111\,\text{N}$ The right cable is at its limiting tension.	6
Notes
Let the left and right tensions be $L$ and $R$ . Horizontal equilibrium gives $L\cos30^\circ=R\cos60^\circ$ , so $R=\sqrt3L$ . Vertical equilibrium gives $W=L\sin30^\circ+R\sin60^\circ=L/2+(\sqrt3L)(\sqrt3/2)=2L$ . The left-cable constraint gives $W\leq160\,\text{N}$ . The right-cable limit gives $R=96$ , so $L=96/\sqrt3$ and $W=192/\sqrt3=64\sqrt3=110.851\ldots\,\text{N}$ . This is the lower limit, so the greatest possible weight is $111\,\text{N}$ and the right cable is limiting.

(6 marks)

Tier 3 · Hard

Three horizontal tugs act on a floating platform in equilibrium. One tug has magnitude

14\,\text{N}

and bearing

030^\circ

; a second has magnitude

20\,\text{N}

and bearing

150^\circ

. Find the exact magnitude of the third tug and its bearing to the nearest degree.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Magnitude $=2\sqrt{79}\,\text{N}$ Bearing $287^\circ$	6
Notes
Take east and north as the positive component directions. The first two forces have total east component $14\sin30^\circ+20\sin150^\circ=17$ and total north component $14\cos30^\circ+20\cos150^\circ=-3\sqrt3$ . For equilibrium, the third force must therefore have components $(-17,3\sqrt3)$ . Its magnitude is $\sqrt{17^2+(3\sqrt3)^2}=\sqrt{316}=2\sqrt{79}\,\text{N}$ . It points west and north. Its angle west of north is $\arctan(17/(3\sqrt3))=73.0039\ldots^\circ$ , so its bearing is $360^\circ-73.0039\ldots^\circ=286.996\ldots^\circ$ , which rounds to $287^\circ$ .

(6 marks)

Q10

Tier 3 · Hard

10.

During an

18

-second test, a sensor reading is modelled by

S=70+30\cos\left(\dfrac{\pi(t-1)}{4}\right)

for

0\leq t\leq18

, where

t

is measured in seconds. Find every time interval for which

S\geq85

, and find the total time for which this condition holds.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$0\leq t\leq7/3$ , $23/3\leq t\leq31/3$ , and $47/3\leq t\leq18$ Total time $=22/3$ seconds	6
Notes
The condition is $\cos u\geq1/2$ , where $u=\pi(t-1)/4$ . The time interval gives $-\pi/4\leq u\leq17\pi/4$ . In this transformed interval, $\cos u\geq1/2$ on $[-\pi/4,\pi/3]$ , $[5\pi/3,7\pi/3]$ and $[11\pi/3,17\pi/4]$ . Since $t=1+4u/\pi$ , these become the three stated time intervals. Their lengths are $7/3$ , $8/3$ and $7/3$ seconds, giving a total of $22/3$ seconds.

(6 marks)

Verified exam appearances

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