Resultant Forces (Resolving vs. Cosine Rule/Lami's Theorem) and Bearings

~50 min · WME01 · 2.1

WME01 · 2.1 · 50 min

Two forces meeting at a point have exactly one resultant, and this whole topic is really two different ways of finding it — resolve into i/j components and use Pythagoras, or draw the triangle the two forces make and hit it with the cosine rule (or Lami's theorem) directly — plus one conversion, bearing to component, that the exam tests far more literally than it looks: get sin and cos the wrong way round and the wrong answer isn't vague, it's a specific number an examiner report has watched real candidates hand in.

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Forces as vectors: magnitude, direction, and what "resultant" means

Spec 2.1 states it plainly: "Magnitude and direction of a vector. Resultant of vectors may also be required," with the guidance that "students may be required to resolve a vector into two components or use a vector diagram," and that "questions may be set involving the unit vectors i and j." A force is a vector — it has a size (magnitude, in newtons) and a direction — and when more than one force acts at the same point, the single force that would have exactly the same effect as all of them acting together is called the resultant. Finding it is the entire content of this lesson.

The two building blocks are magnitude/direction form and component form, and every technique in this topic is a way of converting between them. Magnitude/direction form states a force as a size and an angle — 8 N due north, or 5 N on a bearing of 070°. Component form writes the same force as $p\mathbf{i} + q\mathbf{j}$ , where $\mathbf{i}$ and $\mathbf{j}$ are unit vectors (magnitude exactly 1) pointing east and north respectively, so $p$ is how much of the force acts due east and $q$ is how much acts due north. Both forms describe the same force exactly; neither is more correct, and the whole skill of this topic is choosing whichever form the next step of a calculation actually needs.

Two forces given in component form add exactly the way their labels suggest: add the i-parts together and the j-parts together, separately. $(p_1\mathbf{i}+q_1\mathbf{j}) + (p_2\mathbf{i}+q_2\mathbf{j}) = (p_1+p_2)\mathbf{i} + (q_1+q_2)\mathbf{j}$ . This is the ENTIRE justification for the resolve-and-Pythagoras method below — component addition is simple only because $\mathbf{i}$ and $\mathbf{j}$ point in fixed, perpendicular directions, so adding forces given at arbitrary angles first requires converting each one into that common i/j frame.

Two routes to the same resultant, and a third worth knowing

Given two forces at an angle to each other, spec 2.1's own two routes are (1) resolve each force into i/j components, add the components, then use Pythagoras and $\arctan$ to turn the resultant's own components back into magnitude and direction; or (2) draw the two forces head-to-tail as two sides of a triangle and apply the cosine rule directly to the third side, which represents the resultant. Both are entirely legitimate — the mark scheme awards full credit for either — and a real WME01 examiner report on exactly this comparison found that "those candidates who resolved in two perpendicular directions and then used Pythagoras tended to have more success than those who attempted to use the cosine rule on a triangle of forces" (Jun 2024, Q2). The method-comparison block below works one scenario both ways so you can see precisely where the cosine-rule route's extra risk comes from.

A genuine link worth knowing: the cosine rule, $a^2 = b^2 + c^2 - 2bc\cos A$ , is not printed anywhere in the M1 section of the exam formula booklet — the booklet's ENTIRE M1 entry reads "there are no formulae given for M1 in addition to those candidates are expected to know," followed by "candidates sitting M1 may also require those formulae listed under Pure Mathematics P1 and P2." The cosine rule lives in the P1/P2 section and is extended to M1 candidates through that one sentence — the same formula, the same booklet page, whether the triangle in front of you came from a trigonometry question or a triangle of forces. Pythagoras, by contrast, appears in neither section: it is assumed prior knowledge that neither booklet needs to supply.

A third route exists for the specific case where three forces are in equilibrium (their resultant is zero) — which includes the two-force case, because P, Q, and the equilibrant (the force $-\mathbf{R}$ , equal and opposite to their resultant) are automatically three concurrent forces in equilibrium. Lami's theorem: for three coplanar, concurrent forces $P$ , $Q$ , $T$ in equilibrium, $\dfrac{P}{\sin\alpha} = \dfrac{Q}{\sin\beta} = \dfrac{T}{\sin\gamma}$ , where $\alpha$ is the angle BETWEEN the other two forces ( $Q$ and $T$ ), and correspondingly for $\beta$ and $\gamma$ . It is not named anywhere in the WME01 spec's own wording — spec 5.1 says only "forces treated as vectors, resolution of forces" — but two independent examiner reports treat it as a fully credited alternative: "another direct method involved applying Lami's theorem" (Jan 2022, Q1), and separately, "an alternative method was to consider a triangle of forces and use the sine and/or cosine rule (or Lami's Theorem)" (Jan 2023, Q6). It is genuinely faster once the three angles are correctly identified, and genuinely more exposed to the same angle-identification risk the cosine-rule route carries — it needs the angle BETWEEN each pair of forces, which for a translated triangle-of-forces diagram is once again not always the angle a quick glance suggests.

Mechanism

Why a bearing turns into sin·i + cos·j — and why that feels backwards

A bearing is a direction measured clockwise from north, given as three figures (000° to 360°) — 070° means "start facing north, turn 70° clockwise." This is a genuinely different convention from a standard angle measured anticlockwise from the positive x-axis (east), which is what most trigonometry questions use and what the arithmetic muscle memory from Pure Mathematics is trained on — and the exam relies on that mismatch. Picture the unit circle with north at the top (the j-axis) and east on the right (the i-axis). Starting due north and rotating clockwise through an angle θ, the point traces out east-component $\sin\theta$ and north-component $\cos\theta$ — sine pairs with east, cosine pairs with north, which is the OPPOSITE pairing from the standard-angle convention (where cosine pairs with the x-axis and sine with the y-axis). The reason is simply which axis the angle is measured FROM: a standard angle starts at the x-axis, so cosine (adjacent to that starting axis) goes with x; a bearing starts at the y-axis (north), so cosine goes with y instead, and sine — the "how far around" function — ends up paired with x. Nothing about sin and cos themselves has changed; only which axis the angle begins its sweep from has. A real WME01 examiner report on exactly this conversion records the cost of missing it: on a bearing-to-vector question, "a significant number showed a lack of knowledge of bearings and an answer of 12i + 16j was often seen" — the correct conversion required swapping the components (Jun 2024, Q7). The marked-solution block below reconstructs this exact error with real, checkable numbers. The one habit that catches it every time: before resolving anything, sketch the bearing as an arrow on a compass rose and ask whether the arrow leans more toward east or more toward north — the LARGER trig value belongs on whichever axis the arrow leans closer to, and a swapped answer fails that sanity check immediately.

Diagram — Two forces at O — P due north, Q on bearing 070° — resolved, and as a triangle of forces

x-axis: East (i) · y-axis: North (j)

OA — force P, 8 N, bearing 000° (due north): Drawn from the common point of application O straight up the j-axis, since a bearing of 000° has no east component at all: 8sin0° = 0, 8cos0° = 8.
AB — force Q, 5 N, bearing 070°, translated to start at A: Q is drawn with the SAME direction as if it started at O (bearing 070° throughout — translating a vector never rotates it) but its tail is moved to sit at A, the head of P. This is the head-to-tail construction that turns two separate forces into one triangle: OA, then AB, then back to O along the third side.
OB — the resultant R, closing the triangle: The one side of the triangle that was not drawn as a force — it is what's left over once P and Q are laid head-to-tail, and it represents their combined effect exactly. Its magnitude is found either by Pythagoras on P and Q's added components, or by the cosine rule applied to triangle OAB.

O — common point of application: Where both real forces P and Q actually act. The triangle OAB is a calculation device built on top of this point, not a picture of where anything physically is.
Angle AOB = 70° — the angle BETWEEN P and Q as drawn from O: This is the angle a question states or a diagram marks directly (070° − 000°), and it is the angle that goes straight into the resolve-and-components method with no adjustment.
Angle OAB = 110° — the angle the COSINE RULE actually needs: The interior angle of the triangle at A, between side AO (P reversed) and side AB (Q). It is the supplement of the 70° angle above, 180° − 70°, because translating Q to start at A does not rotate it — one arm of the original 70° angle is effectively reversed at this vertex.
Bearing of R ≈ 026°: Found from arctan(east-component ÷ north-component) = arctan(4.70 ÷ 9.71) ≈ 25.8°, measured from north — the same bearing convention the mechanism block above derives, applied here to the ANSWER rather than to one of the given forces.

Common error: Applying the cosine rule to the triangle of forces using the 70° angle between P and Q (the angle as drawn from O), instead of its supplement.

Correct: The angle inside the head-to-tail triangle, at the vertex where the two force-vectors meet, is 180° minus the angle between the forces as originally drawn from the common point — here 110°, not 70°. A real examiner report on this exact question type records the cost of the mix-up directly.

examiner-report · Jun 2024 · Q2

Same question, every valid method

Two forces act at a point O: P, of magnitude 8 N, acts due north; Q, of magnitude 5 N, acts on a bearing of 070°. Find the magnitude of the resultant force R, and its bearing. — VERIDIAN-original question (spec 2.1–2.2). The real Jun 2024 Q2 question this is built to mirror gives neither its actual force magnitudes nor its actual angle in the research bank reviewed for this lesson — only the examiner report's commentary on HOW candidates went wrong survives. These numbers (8 N, 5 N, 70°) were chosen to reproduce that exact wrong-angle trap with values that check cleanly both ways, not copied from any real paper.

2 valid methods · every one reaches R ≈ 10.8 N (3 s.f.), on a bearing of 026° · 5 marks available

01
$P = 8\mathbf{j}$ (bearing 000°: no east component). $Q$ : east component $= 5\sin70° = 4.70$ , north component $= 5\cos70° = 1.71$ , so $Q = 4.70\mathbf{i} + 1.71\mathbf{j}$ .
Method mark for resolving BOTH forces into i/j components using the bearing convention (sin → east, cos → north). Every term needing resolving must actually be multiplied by sin or cos for this mark — an unresolved force copied straight across does not qualify.
M1
02
$\mathbf{R} = P + Q = 4.70\mathbf{i} + (8+1.71)\mathbf{j} = 4.70\mathbf{i} + 9.71\mathbf{j}$
Accuracy mark for the resultant's own component form, both components correct.
A1
03
$R = \sqrt{4.70^2 + 9.71^2} = \sqrt{22.09 + 94.28} = \sqrt{116.4} = 10.8$ N (3 s.f.)
Accuracy mark for the magnitude, by Pythagoras on the resultant's own components — no angle-identification step is needed here at all, because Pythagoras only ever uses the right angle between the i- and j-axes themselves.
A1
04
$\arctan\left(\dfrac{4.70}{9.71}\right) = 25.8°$ , so the bearing of R is $\mathbf{026°}$ (to the nearest degree, written as a three-figure bearing).
Accuracy mark for the bearing, read off using the SAME sin-goes-with-east convention as the mechanism block above, now applied in reverse to recover an angle from known components.
A1

This is the route a real Jun 2024 examiner report says had "more success" — and the reason is structural, not just habit: Pythagoras never needs an angle identified at all, because the i- and j-axes are always at exactly 90° to each other by definition. Every risk in this method is contained inside the resolving step, and resolving is spec 2.1's own explicitly named technique ("students may be required to resolve a vector into two components"), so there is no second, less-familiar formula being reached for under pressure.

In your own words

In one sentence: why is the angle used in the triangle-of-forces cosine rule the SUPPLEMENT of the angle between the two forces, rather than that angle itself?

Marked, line by line

A force F, of magnitude 20 N, acts on a bearing of 053°. Express F in the form $p\mathbf{i} + q\mathbf{j}$ (N), where $\mathbf{i}$ and $\mathbf{j}$ are unit vectors due east and due north respectively, giving each component to 3 s.f. (2 marks) — VERIDIAN-original question. Its own figures are not from a real WME01 paper: the research bank for this lesson verifies only the WRONG ANSWER a real Jun 2024 Q7 examiner report records, "12i + 16j," not the real question's actual magnitude or bearing. The 20 N and 053° used here were chosen specifically so the correct working reproduces that exact reported wrong answer once sin and cos are swapped — 20sin53° = 16.0 (3 s.f.) and 20cos53° = 12.0 (3 s.f.), and 12² + 16² = 400 = 20², so 20 N is the magnitude those two reported numbers themselves imply.

2 marks available

01
East (i) component $= F\sin(\text{bearing}) = 20\sin53° = 15.97...$ N. North (j) component $= F\cos(\text{bearing}) = 20\cos53° = 12.04...$ N.
Method mark for resolving using the bearing convention correctly matched to each axis — sine multiplied against the EAST component, cosine against the NORTH component, because the bearing is measured from north, not from east.
M1
02
$\mathbf{F} \approx 16.0\mathbf{i} + 12.0\mathbf{j}$ (N), each component to 3 s.f.
Accuracy mark for both components correct AND in the correct slot. A component with the right two numbers (16.0 and 12.0) but swapped between i and j would not earn this mark — the slot each number sits in is exactly what the mark is checking.
A1

Complete it yourself

Complete the chain — express a force of 10 N on a bearing of 120° as $p\mathbf{i} + q\mathbf{j}$ (N)

01
Sketch first. A bearing of 120° is between due east (090°) and due south (180°), so the force should have a positive east component and a NEGATIVE north component (some of its effect is to the south). Any final answer with a positive j-component would already be inconsistent with this sketch, before a single number is computed.
02
Apply the bearing convention from the mechanism block: for a force of magnitude $F$ on bearing $\theta$ , east/i-component $= F\sin\theta$ , north/j-component $= F\cos\theta$ . Here $F = 10$ , $\theta = 120°$ .

Named traps

bearing-sin-cos-pairing-swapped: Confirmed directly on a real bearing-to-vector question: "a significant number showed a lack of knowledge of bearings and an answer of 12i + 16j was often seen" — the report is explicit that "the correct conversion required swapping the components" (Jun 2024, Q7). The mechanism is treating a bearing (measured clockwise from NORTH) like a standard angle (measured anticlockwise from EAST), which swaps which axis sine belongs to. The marked-solution block above reconstructs this exact wrong answer with real, checkable numbers.
triangle-of-forces-supplement-not-taken: Confirmed on a real resultant-of-two-forces question that directly compared the two standard methods: "those candidates who resolved in two perpendicular directions and then used Pythagoras tended to have more success than those who attempted to use the cosine rule on a triangle of forces. A common error with this approach was to use an incorrect angle of 30, 60 or in some cases 330 degrees, even when they had drawn what appeared to be the correct obtuse angled triangle" (Jun 2024, Q2). The triangle itself being right and the angle plugged into the formula being wrong is precisely the failure mode a correct sketch does not, by itself, protect against — the interior angle at the shared vertex is the SUPPLEMENT of the angle between the forces as drawn from their common point, never that angle directly.
incomplete-resolution-missing-a-term: Confirmed on a real equilibrium/resolving question, where a candidate resolved in one direction but dropped a whole force from the equation: "incomplete resolutions e.g. 5 = Fcos30 rather than 5 = Fcos30 + Tcos60" (Jan 2022, Q1, the same series and question that independently confirms Lami's theorem as a credited alternative method — "another direct method involved applying Lami's theorem"). This is a method error, not an accuracy error, under the mark scheme's own general principles for Mechanics marking: an equation produced by resolving is only creditable once every force with a component in that direction has actually been multiplied by its own sin or cos and included — one omitted term breaks the method mark, whatever the rest of the arithmetic does afterwards.

Retrieval — with feedback on every choice

Question 1

2 marks

A force of magnitude 12 N acts on a bearing of 200°. Which pair of components, to 3 s.f., correctly expresses it as $p\mathbf{i} + q\mathbf{j}$ (N)?

Question 2

3 marks

Two forces of magnitude 9 N and 4 N act at a point, with an angle of 100° between them as drawn from their common point. Using the cosine rule on the triangle of forces, what is the magnitude of the resultant?

Question 3

2 marks

A real WME01 examiner report directly compared two methods on the same resultant-of-two-forces question and found one had "more success" than the other. Which method, and why is that consistent with the mechanics of the two approaches rather than just a coincidence of that one series?

Question 4

1 mark

Which of these is a legitimate reason to reach for Lami's theorem instead of resolving into components, based on how real WME01 examiner reports describe it?

Reference — not a study method, a lookup

Bearing → components (i = east, j = north): i-part = F sinθ, j-part = F cosθ. Sine goes with EAST — the opposite pairing from a standard angle.
Resultant of P, Q at angle θ (from a common point): R² = P² + Q² + 2PQcosθ — same as resolve-then-Pythagoras, always.
Triangle of forces (head-to-tail): the cosine-rule angle at the shared vertex is 180° − θ, the SUPPLEMENT of the angle between the forces at their common point.
Lami's theorem (3 concurrent forces in equilibrium): P/sinα = Q/sinβ = T/sinγ, each angle opposite the OTHER two forces.
Cosine rule is a P1/P2 formula extended to M1 via the booklet's own wording — not printed again in the M1 section. Pythagoras is assumed knowledge, printed nowhere.

Not affiliated with or endorsed by Pearson Edexcel. Every direct quotation attributed to a mark scheme or examiner report in this lesson (Jun 2024 Q2, Jun 2024 Q7, Jan 2022 Q1, Jan 2023 Q6) is copied verbatim from WME01-verified-facts.md, itself checked against the primary Pearson PDF. The research bank verifies the ERROR PATTERNS on all three of these — a specific wrong answer (12i + 16j), a specific class of wrong angle (30°, 60°, 330°), and Lami's theorem's status as a credited alternative — but not the real questions' own force magnitudes, angles, or bearings, which were not available in the source material reviewed for this pass. Every worked scenario in this lesson carrying actual numbers (8 N due north / 5 N on bearing 070°; 20 N on bearing 053°; 10 N on bearing 120°; 12 N on bearing 200°; 9 N and 4 N at 100°) is therefore VERIDIAN-original, built specifically to reproduce the verified error patterns exactly rather than to gesture at them loosely, and is not a reproduction of any real Pearson question. All arithmetic was checked independently before being written in.

Question 12 marks

A force of magnitude 12 N acts on a bearing of 200°. Which pair of components, to 3 s.f., correctly expresses it as $p\mathbf{i} + q\mathbf{j}$ (N)?

$-4.10\mathbf{i} - 11.3\mathbf{j}$ , since $12\sin200° = -4.10$ and $12\cos200° = -11.3$
Correct, and both signs are worth checking against a sketch: 200° is between south (180°) and west (270°), so the force should have a small negative east component and a larger negative north component — both consistent with what the sin/cos values give here.
B $-11.3\mathbf{i} - 4.10\mathbf{j}$ , since $12\cos200° = -11.3$ and $12\sin200° = -4.10$
The sin/cos pairing is swapped — cosine attached to east and sine attached to north, the standard-angle convention rather than the bearing convention. A bearing is measured from north, so sine belongs with east and cosine with north, exactly as the mechanism block in this lesson derives.
C $4.10\mathbf{i} + 11.3\mathbf{j}$
The magnitudes are right and both signs are wrong. A bearing of 200° points into the south-west quadrant, so both components must be negative — a positive answer here has not been checked against a sketch of where 200° actually points.
D $-4.10\mathbf{i} + 11.3\mathbf{j}$
The east component's sign is right; the north component's is not. At 200° the force leans south, not north, so the j-component has to be negative as well — only one of the two signs has been fixed here.

Traps tested: Bearing sin cos pairing swapped · Component sign not checked against sketch

Question 23 marks

$R = \sqrt{9^2+4^2-2(9)(4)\cos80°} \approx 9.19$ N
Correct. The triangle's interior angle is the supplement of 100°, which is 80°, not 100° itself: $R^2 = 81+16-72\cos80° = 97-72(0.174)=97-12.5=84.5$ , so $R \approx 9.19$ N. (Rounding check: $\sqrt{84.5}=9.19$ .) Applying the supplement is the step this whole topic keeps testing.
B $R = \sqrt{9^2+4^2-2(9)(4)\cos100°} \approx 10.5$ N
This plugs the 100° angle BETWEEN the forces directly into the triangle-of-forces cosine rule, without taking its supplement first — exactly the error a real Jun 2024 examiner report records on this question type, where the triangle drawn looked correct but the angle substituted did not match it.
C $R = 9 + 4 = 13$ N
Force magnitudes only add like scalars when the two forces act in exactly the same direction. At 100° apart, part of each force works against the other's direction, so the resultant is shorter than the simple sum.
D $R = \sqrt{9^2-4^2} \approx 8.06$ N
Pythagoras only applies when the two forces are at exactly 90° to each other, and even then it is $\sqrt{9^2+4^2}$ (added, not subtracted) unless one force is genuinely the resultant's own component. At 100°, neither shortcut applies — the cosine rule (or resolving into components) is required.

Traps tested: Triangle of forces supplement not taken · Vector magnitudes added as scalars · Pythagoras misapplied to non right angle

Question 32 marks

Resolving into components then Pythagoras — because Pythagoras needs no angle to be identified at all, while the cosine-rule route needs the triangle's interior angle correctly related to the angle between the forces
Correct. The report's own words: "those candidates who resolved in two perpendicular directions and then used Pythagoras tended to have more success than those who attempted to use the cosine rule on a triangle of forces" (Jun 2024, Q2). The structural reason survives beyond that one series: the i/j axes are always exactly perpendicular by definition, so Pythagoras never has an angle-identification step to get wrong, whereas the triangle-of-forces angle has to be derived (as the supplement) every single time.
BThe cosine rule on a triangle of forces — because it needs fewer steps than resolving into components first
This reverses the report's actual finding. The cosine-rule route can be shorter on the page, but shorter is not the same as safer: the step it saves (resolving) is exactly the step that has no angle-identification risk, while the step it keeps (the interior angle) is exactly where the report records candidates going wrong.
CBoth methods scored identically in that series, so no comparison can genuinely be drawn
The report makes an explicit, direct comparison rather than reporting equal outcomes — this is a real, stated finding, not a null result being over-read.
DLami's theorem — because it is the fastest of the three methods available on this question type
The report's comparison is specifically between resolving-and-Pythagoras and the cosine rule; Lami's theorem is confirmed elsewhere in the archive as a legitimate alternative (Jan 2022, Jan 2023) but was not the subject of this particular comparison, and it shares the cosine-rule route's own angle-identification risk rather than avoiding it.

Traps tested: Shorter method assumed safer · Overclaims uncertainty · Unsupported method substituted into a specific finding

Question 41 mark

Which of these is a legitimate reason to reach for Lami's theorem instead of resolving into components, based on how real WME01 examiner reports describe it?

It is confirmed, in two independent series, as a directly credited alternative method for a system of concurrent forces in equilibrium — worth knowing as a genuine option, not a shortcut invented outside the mark scheme
Correct. "Another direct method involved applying Lami's theorem" (Jan 2022, Q1) and, independently, "an alternative method was to consider a triangle of forces and use the sine and/or cosine rule (or Lami's Theorem)" (Jan 2023, Q6) — two separate series both treating it as legitimate, examiner-credited working, even though the spec's own wording never names it.
BIt is printed in the M1 section of the exam formula booklet, so it can be looked up rather than recalled
The M1 section of the booklet has no formulae at all beyond what it extends from P1/P2 — and Lami's theorem is not among those extended formulae either. Like the resolve-and-Pythagoras route's underlying trig, it has to be recalled or derived, not looked up.
CIt removes the need to identify any angle correctly, unlike the cosine-rule route
The opposite is true: Lami's theorem needs the angle BETWEEN each pair of the three concurrent forces, which for a translated triangle-of-forces diagram carries exactly the same supplement risk the cosine-rule route has — it is not a way of avoiding that risk, only a different formula built on the same triangle.
DIt works for any two forces, whether or not a third force or an equilibrant is involved
Lami's theorem is stated for three concurrent forces in equilibrium. Two forces alone need a third — the equilibrant, equal and opposite to their resultant — brought into the picture before the theorem applies at all.

Traps tested: Formula booklet content overclaimed · Lamis theorem assumed angle free · Lamis theorem applied to two forces only

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jun 2024 · Q2 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WME01.

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Every one of the five suvat equations assumes exactly one thing: that the acceleration is the same constant value for the whole interval you apply it to. M1's single richest, most consistently examined trap is a journey that quietly breaks that assumption partway through — a force removed, a surface that roughens, a parachute that opens — while the equation on the page looks exactly as innocent as it did a line before. Independently confirmed in at least two separate series, this is the one habit that decides whether a two-stage journey earns full marks or, in the examiners' own words, "no credit" at all.

55 min