Edexcel International A Level Maths · P1 · Pure Mathematics 1
Trigonometry: triangles, radians and graphs
28 free, high-quality past-paper-style questions on Trigonometry: triangles, radians and graphs for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 239 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P1-5-B-01P1 · ASEasy7 marks
A triangular sail PQR has PQ = 5.4 m, angle QPR = 52° and angle PQR = 61°.
Figure: the sail PQR (not to scale)
(a)
Find the length of QR, giving your answer to 3 significant figures.
(3)
(b)
Find the length of PR.
(2)
(c)
Find the area of the sail.
(2)
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Answer
Mark
Notes
Part (a)
angle PRQ = 180° − 52° − 61° = 67°
B1
QR/sin 52° = 5.4/sin 67°
M1
QR = 4.62 m (4.6227…)
A1
Part (b)
PR = 5.4 sin 61°/sin 67°
M1
= 5.13 m (5.1308…)
A1
Part (c)
½ × 5.4 × 5.1308 × sin 52° (or any correct ½ab sin C with the included angle)
M1
= 10.9 m2
A1
Exam tip: Find the third angle first: with all three angles known, every side can be reached from the one complete pair (PQ with the 67° angle).
Write down the coordinates of the point where C meets the y-axis.
(1)
(b)
Find the exact coordinates of the points where C meets the x-axis.
(2)
(c)
Write down the coordinates of the minimum point of C.
(1)
(d)
State the number of solutions of the equation 4 cos x = 1 in the interval 0 ≤ x ≤ 2π.
(1)
(e)
Given that 4 cos 0.4 = 3.684 to 4 significant figures, use the symmetry of C to write down another value of x in the interval 0 ≤ x ≤ 2π for which 4 cos x = 3.684, giving your answer to 3 decimal places.
(1)
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Answer
Mark
Notes
Part (a)
(0, 4)
B1
Part (b)
(π/2, 0)
B1
(3π/2, 0)
B1
Part (c)
(π, −4)
B1
Part (d)
2 (the line y = 1 crosses C twice)
B1
Part (e)
x = 2π − 0.4 = 5.883
B1
Model sketch
Figure: model sketch of C with the line y = 1
Exam tip: The cosine curve is symmetrical about x = π on 0 ≤ x ≤ 2π, so if α is one solution, 2π − α is another.
Figure 1 shows a triangular flower bed PQR. The lengths of its three sides are marked on the figure.
(a)
Explain which angle of the triangle is the largest.
(1)
(b)
Find the size of angle PQR, giving your answer to 1 decimal place.
(3)
(c)
Find the area of the flower bed, giving your answer to 3 significant figures.
(2)
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Answer
Mark
Notes
Part (a)
the largest angle is opposite the longest side, PR = 11.0 m, so it is angle PQR
B1
Part (b)
cos Q = (5.82 + 7.42 − 11.02)/(2 × 5.8 × 7.4)
M1
= −32.6/85.84 = −0.3798
A1
angle PQR = 112.3° (112.32…)
A1
Part (c)
½ × 5.8 × 7.4 × sin 112.3°
M1
= 19.9 m2 (19.85…)
A1
Common error: Dropping the minus sign. A negative cosine means an obtuse angle, and cos−1(−0.3798) gives 112.3° directly — there is no need to subtract from 180°.
Figure 1 shows the region of lawn watered by a rotating sprinkler at O. The region is a sector OAB of a circle; its radius and the angle at O are marked on the figure.
(a)
Write angle AOB in radians, as an exact multiple of π.
(1)
(b)
Find the exact length of the arc AB.
(2)
(c)
Find the area of lawn watered, giving your answer to 3 significant figures.
(2)
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Answer
Mark
Notes
Part (a)
135 × π/180 = 3π/4
B1
Part (b)
s = rθ = 6 × 3π/4
M1
= 9π/2 m (or 4.5π m)
A1
Part (c)
½ × 62 × 3π/4 (= 27π/2)
M1
= 42.4 m2 (42.41…)
A1
Exam tip: Convert an angle given in degrees to radians before using s = rθ or ½r2θ: multiply by π/180, and keep π in the answer when an exact value is asked for.
Figure 1 shows the curve with equation y = a cos bx, 0 ≤ x ≤ π, where a and b are positive constants.
(a)
Use the figure to write down the value of a.
(1)
(b)
Use the figure to state the period of the curve, and hence find the value of b.
(2)
(c)
Find the exact coordinates of the points where the curve meets the x-axis.
(2)
(d)
State the number of solutions of the equation a cos bx = 1 for 0 ≤ x ≤ π.
(1)
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Answer
Mark
Notes
Part (a)
a = 2 (the greatest value of y)
B1
Part (b)
period = 2π/3 (consecutive maxima at x = 0 and x = 2π/3)
B1
2π/b = 2π/3, so b = 3
A1
Part (c)
3x = π/2, 3π/2, 5π/2: (π/6, 0) and (π/2, 0)
B1
(5π/6, 0)
B1
Part (d)
3 (the line y = 1 crosses the curve three times)
B1
Exam tip: The period of cos bx is 2π/b. Read the period from the graph as the distance between two neighbouring maxima (or minima), then solve 2π/b = period.
Figure 1 shows two coastguard stations A and B, where B is 15 km due east of A. A yacht at C is on a bearing of 052° from A and on a bearing of 310° from B. Treat A, B and C as points in a horizontal plane.
(a)
Show that angle CAB = 38° and angle CBA = 40°.
(2)
(b)
Find the distance AC.
(3)
(c)
Find the shortest distance from the yacht to the straight line AB.
(2)
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Answer
Mark
Notes
Part (a)
angle CAB = 90° − 52° = 38°
B1
angle CBA = 310° − 270° = 40°
B1
Part (b)
angle ACB = 180° − 38° − 40° = 102°
B1
AC/sin 40° = 15/sin 102°
M1
AC = 9.86 km (9.857…)
A1
Part (c)
AC sin 38° (or BC sin 40° with BC = 9.441)
M1
= 6.07 km (6.068…)
A1
Exam tip: Bearings are measured clockwise from north, so due east is 090° and due west is 270°. The angle a bearing makes with an east–west line comes from 90° or 270°.
In triangle ABC, AB = 11 cm, AC = 8 cm and angle ABC = 35°.
Figure: the two possible positions C₁ and C₂ of C (not to scale)
(a)
Show that sin(angle ACB) = 0.7887 to 4 decimal places.
(2)
(b)
Find the two possible sizes of angle ACB, giving your answers to 1 decimal place.
(2)
(c)
Given that angle ACB is obtuse, find the length of BC.
(3)
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Answer
Mark
Notes
Part (a)
sin C/11 = sin 35°/8
M1
sin C = 11 sin 35°/8 = 0.78867… = 0.7887
A1
Part (b)
52.1°
A1
180° − 52.1° = 127.9° (ft)
B1
Part (c)
angle BAC = 180° − 35° − 127.9° = 17.1°
M1
BC = 8 sin 17.1°/sin 35°
M1
BC = 4.09 cm (4.10 from the rounded angle is acceptable)
A1
Common error: Giving only the calculator value 52.1°. The given angle is opposite the shorter given side (8 < 11), so the obtuse angle 180° − 52.1° may also fit — check the angle sum.
A survey drone takes off from a point L. It flies 3.2 km on a bearing of 040° to a point X, then 5.0 km on a bearing of 125° to a point Y. Treat the drone as moving in a horizontal plane.
Figure: the path of the drone (not to scale)
(a)
Show that angle LXY = 95°.
(2)
(b)
Find the distance LY.
(3)
(c)
Find the bearing of Y from L, giving your answer to the nearest 0.1°.
(3)
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Answer
Mark
Notes
Part (a)
bearing of L from X = 040° + 180° = 220°
M1
angle LXY = 220° − 125° = 95°
A1
Part (b)
LY2 = 3.22 + 5.02 − 2 × 3.2 × 5.0 × cos 95°
M1
LY2 = 38.03
A1
LY = 6.17 km
A1
Part (c)
sin XLY = 5.0 sin 95°/6.1667
M1
angle XLY = 53.9°
A1
bearing = 040° + 53.9° = 093.9°
A1
Exam tip: Draw a north line at each turning point. The angle inside the triangle at X comes from the back bearing (220°) and the new bearing (125°).
A circular tabletop has centre O and radius 60 cm. To fit it against a wall, a straight cut is made along a chord AB whose perpendicular distance from O is 40 cm. The smaller piece is removed. Angle AOB = θ radians.
Figure: the tabletop with the removed piece shaded (not to scale)
(a)
Show that θ = 1.682 to 3 decimal places.
(2)
(b)
Find the area of the piece that is removed.
(3)
(c)
Find the percentage of the area of the original tabletop that is removed.
(2)
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Answer
Mark
Notes
Part (a)
cos(θ/2) = 40/60
M1
θ = 2 cos−1(2/3) = 1.68213… = 1.682
A1
Part (b)
sector: ½ × 602 × 1.6821 (= 3027.8)
M1
triangle: ½ × 602 × sin 1.6821 (= 1788.9)
M1
segment = 1240 cm2 (1239.0)
A1
Part (c)
1239.0 ÷ (π × 602) × 100
M1
= 11.0%
A1
Exam tip: Segment = sector − triangle. Work in radians throughout, including sin θ in the triangle area: ½r2(θ − sin θ).
A windscreen wiper arm turns about a fixed point O. The rubber blade lies along the arm, from 18 cm to 52 cm from O. The arm turns through an angle of 1.95 radians, and the blade wipes the region between the two circular arcs of radii 18 cm and 52 cm centred at O.
Figure: the region wiped by the blade (shaded; not to scale)
(a)
Find the length of the arc traced out by the outer end of the blade.
(2)
(b)
Find the area of the region wiped by the blade.
(3)
(c)
Find the perimeter of the region wiped by the blade.
(2)
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Answer
Mark
Notes
Part (a)
52 × 1.95
M1
= 101.4 cm
A1
Part (b)
½ × 522 × 1.95 (= 2636.4)
M1
− ½ × 182 × 1.95 (= 315.9)
M1
= 2320 cm2 (2320.5)
A1
Part (c)
101.4 + 18 × 1.95 + 2 × 34
M1
= 204.5 cm
A1
Exam tip: The perimeter of this shape has four pieces: two arcs and two straight edges, each 52 − 18 = 34 cm long.
The curve C has equation y = tan(x − π/6), 0 ≤ x ≤ 2π.
(a)
State the period of the graph of y = tan x.
(1)
(b)
Find the equations of the asymptotes of C.
(2)
(c)
Find the exact coordinates of the points where C meets the x-axis.
(2)
(d)
Find the exact y-coordinate of the point where C meets the y-axis.
(2)
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Answer
Mark
Notes
Part (a)
π (or 180°)
B1
Part (b)
x − π/6 = π/2 gives x = 2π/3
B1
x − π/6 = 3π/2 gives x = 5π/3
B1
Part (c)
(π/6, 0)
B1
(7π/6, 0)
B1
Part (d)
y = tan(−π/6) = −tan(π/6)
M1
= −√3/3 (or −1/√3)
A1
Model sketch
Figure: model sketch of C with its asymptotes
Exam tip: The asymptotes of tan are where the inside of the bracket equals π/2 + nπ. Solve “inside = π/2, 3π/2, …” and keep only the values in the interval.
Figure 1 shows part of a mechanism. A straight rail starts at P and makes an angle of 40° with the line PQ, where PQ = 10 cm. A rod of length 7 cm is pivoted at Q, and its other end R rests on the rail. There are two possible positions, R1 and R2, as shown. Model the rod and the rail as straight lines in one plane.
(a)
Show that, for either position, sin(angle PRQ) = 0.9183 to 4 decimal places.
(2)
(b)
Find the size of angle PR1Q and the size of angle PR2Q, giving your answers to 1 decimal place.
(2)
(c)
Find the distances PR1 and PR2.
(4)
(d)
Find the distance between the two possible positions of R.
(or the cosine rule: x2 − 20 cos 40° x + 51 = 0 with both roots)
—
Part (d)
10.43 − 4.89 = 5.54 cm (ft their (c))
B1
Common error: Stopping at the calculator value 66.7°. The side opposite the 40° angle (7 cm) is shorter than PQ (10 cm), so 180° − 66.7° also fits and gives the second position of the rod.
Figure 1 shows a circle with centre O and radius 6 cm. The point T is 10 cm from O. The lines TA and TB are tangents to the circle at A and B. The shaded region is bounded by TA, TB and the minor arc AB.
(a)
Explain why angle OAT is a right angle, and find the length of TA.
(2)
(b)
Show that angle AOB = 1.855 radians, to 3 decimal places.
(2)
(c)
Find the area of the shaded region.
(3)
(d)
Find the perimeter of the shaded region.
(2)
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Answer
Mark
Notes
Part (a)
a tangent is perpendicular to the radius at the point of contact
B1
TA = √(102 − 62) = 8 cm
B1
Part (b)
cos(angle AOT) = 6/10, so angle AOT = 0.9273
M1
angle AOB = 2 × 0.92730 = 1.85459… = 1.855
A1
Part (c)
area of OATB = 2 × ½ × 6 × 8 = 48
B1
sector OAB = ½ × 62 × 1.8546 = 33.38
M1
shaded area = 48 − 33.38 = 14.6 cm2 (14.62)
A1
Part (d)
arc AB = 6 × 1.8546 = 11.13
M1
perimeter = 8 + 8 + 11.13 = 27.1 cm
A1
Exam tip: The right angle between a tangent and a radius turns OAT into a right-angled triangle, so Pythagoras gives TA and cos−1(r/OT) gives half of the angle at O.
Figure 1 shows the curves C1 with equation y = sin x° and C2 with equation y = cos(x − 30)°, for 0 ≤ x ≤ 360. The two points where the curves cross are marked with dots.
(a)
Describe fully the single transformation that maps the graph of y = cos x° onto C2.
(2)
(b)
Write down the coordinates of the maximum point and of the minimum point of C2.
(2)
(c)
Show that the curves cross at the point where x = 60.
(2)
(d)
Use the figure to write down the x-coordinate of the other crossing point, and verify that both curves have the same y-value there.
(2)
(e)
Find the set of values of x, 0 ≤ x ≤ 360, for which sin x° and cos(x − 30)° are both negative.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
translation
B1
by 30 in the positive x-direction, i.e. by the vector (30, 0)
B1
Part (b)
maximum (30, 1)
B1
minimum (210, −1)
B1
Part (c)
sin 60° = √3/2
M1
cos(60 − 30)° = cos 30° = √3/2, the same value, so the curves meet at x = 60
A1
Part (d)
x = 240
B1
sin 240° = −√3/2 and cos(240 − 30)° = cos 210° = −√3/2
B1
Part (e)
180 < x < 300
B1
Common error: Moving the curve the wrong way. cos(x − 30)° reaches each value 30 later than cos x°, so the maximum moves from x = 0 to x = 30: the graph shifts to the right.
Figure 1 shows a pendulum. A small bob is attached to a light inextensible string of length 1.6 m, whose other end is fixed at O. The bob swings in a vertical plane between its extreme positions A and B, moving along the circular arc AB of length 0.9 m. M is the lowest point of the arc, directly below O. Model the bob as a particle.
(a)
Find angle AOB in radians, and convert it to degrees, giving your answer to 1 decimal place.
(3)
(b)
Find the straight-line distance AB.
(2)
(c)
Find the height of the bob at A above its height at M, giving your answer in centimetres.
(2)
(d)
Find the area swept out by the string as the bob moves from A to B.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
0.9 = 1.6θ
M1
θ = 0.5625 rad
A1
= 0.5625 × 180/π = 32.2°
B1
Part (b)
AB = 2 × 1.6 sin 0.28125 (or the cosine rule with angle 0.5625)
M1
= 0.888 m (0.8882)
A1
Part (c)
1.6 − 1.6 cos 0.28125
M1
= 0.0629 m = 6.29 cm
A1
Part (d)
½ × 1.62 × 0.5625 = 0.72 m2
B1
Exam tip: The vertical OM cuts the swing into two right-angled triangles with angle θ/2 at O. Half the chord is 1.6 sin(θ/2) and the depth of the bob below O is 1.6 cos(θ/2).
Figure 1 shows the curve with equation y = tan 2x, 0 ≤ x ≤ π, together with the line y = 3. The point (π/8, 1) on the curve is marked.
(a)
State the period of tan 2x.
(1)
(b)
Write down the equations of the asymptotes of the curve in the interval shown.
(2)
(c)
Write down the exact coordinates of the points where the curve meets the x-axis.
(2)
(d)
Given that the smallest positive solution of tan 2x = 3 is x = 0.6245 to 4 decimal places, use the period to find the other solution in the interval 0 ≤ x ≤ π, giving your answer to 3 decimal places.
(1)
(e)
Using the marked point and the shape of the curve, write down all the solutions of tan 2x = 1, and all the solutions of tan 2x = −1, in the interval 0 ≤ x ≤ π.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
π/2
B1
Part (b)
x = π/4
B1
x = 3π/4
B1
Part (c)
(0, 0) and (π/2, 0)
B1
(π, 0)
B1
Part (d)
0.6245 + π/2 = 2.195
B1
Part (e)
tan 2x = 1: x = π/8, 5π/8
B1
tan 2x = −1: x = 3π/8, 7π/8
B1
Exam tip: tan kx has period π/k, so once one solution is known the others in the interval are found by adding (or subtracting) π/k. For tan 2x that step is π/2.
Figure 1 shows the design for a flat metal plate. OAB is a sector of a circle with centre O, radius 10 cm and angle 0.8 radians at O. The point C lies on OA, and BC is perpendicular to OA. The plate is the shaded region bounded by the arc AB and the straight lines BC and CA.
(a)
Find the length of BC.
(2)
(b)
Find the length of CA.
(2)
(c)
Find the area of the plate.
(3)
(d)
Find the perimeter of the plate.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
BC = 10 sin 0.8
M1
= 7.17 cm (7.1736)
A1
Part (b)
OC = 10 cos 0.8 = 6.967
M1
CA = 10 − 6.967 = 3.03 cm (3.033)
A1
Part (c)
sector OAB = ½ × 102 × 0.8 = 40
M1
triangle OCB = ½ × 6.967 × 7.174 = 24.99
M1
plate = 40 − 24.99 = 15.0 cm2 (15.01)
A1
Part (d)
10 × 0.8 + 7.17 + 3.03 = 18.2 cm (ft)
B1
Common error: Working out sin 0.8 and cos 0.8 with the calculator in degree mode. The angle is in radians, so set radian mode first: sin 0.8 = 0.717, not 0.0140.
A sector of a circle has radius r cm and angle θ radians at the centre. The perimeter of the sector is 30 cm and its area is 50 cm2.
Figure: a sector of radius r cm and angle θ radians (not to scale)
(a)
Show that r2 − 15r + 50 = 0.
(4)
(b)
Find the two possible pairs of values of r and θ.
(3)
(c)
For the sector with θ < π, find the area of the segment between the arc and the chord joining the ends of the arc.
(3)
(d)
For the same sector, find the length of the chord.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
2r + rθ = 30
B1
½r2θ = 50
B1
eliminating θ, e.g. rθ = 100/r so 2r + 100/r = 30
M1
2r2 − 30r + 100 = 0, so r2 − 15r + 50 = 0
A1
Part (b)
(r − 5)(r − 10) = 0
M1
r = 5 or r = 10
A1
r = 5, θ = 4 and r = 10, θ = 1
A1
Part (c)
triangle area ½ × 102 × sin 1 (= 42.07)
M1
segment = 50 − 42.07
M1
= 7.93 cm2
A1
Part (d)
2 × 10 sin 0.5 = 9.59 cm (or by the cosine rule)
B1
Exam tip: In a “show that”, write the line just before the given result, here 2r2 − 30r + 100 = 0, so that every step to the printed answer can be seen.
A church window has a lower part that is a rectangle ABCD with DC along the bottom, AB = 1.2 m and AD = 1.5 m. Above AB is a pointed arch. The arch is bounded by AB and by two circular arcs that meet at the top point P: arc AP has centre B and radius 1.2 m, and arc BP has centre A and radius 1.2 m.
Figure: the window (not to scale)
(a)
Explain why triangle ABP is equilateral, and hence write down the size of angle ABP in radians.
(2)
(b)
Show that the height of P above DC is (1.5 + 0.6√3) m.
(2)
(c)
Find the exact length of the whole outer edge of the window, giving your answer in the form a + bπ.
(3)
(d)
Show that the area of the arch above AB is (0.48π − 0.36√3) m2.
(3)
(e)
Find the total area of glass in the window, giving your answer to 3 significant figures.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
BP = BA = 1.2 (radii of the arc centred at B) and AP = AB = 1.2 (radii of the arc centred at A), so all three sides are 1.2 m
B1
angle ABP = π/3
B1
Part (b)
height of triangle ABP = 1.2 sin(π/3) = 0.6√3
M1
total = 1.5 + 0.6√3
A1
Part (c)
each arc = 1.2 × π/3 = 0.4π
M1
both arcs = 0.8π
A1
edge = 1.5 + 1.2 + 1.5 + 0.8π = 4.2 + 0.8π m
A1
Part (d)
each sector = ½ × 1.22 × π/3 = 0.24π
M1
the two sectors overlap in triangle ABP, so arch = 2 sectors − triangle
M1
triangle = ½ × 1.22 × sin(π/3) = 0.36√3, so arch = 0.48π − 0.36√3
A1
Part (e)
1.2 × 1.5 + 0.48π − 0.36√3
M1
= 2.68 m2
A1
Exam tip: When two sectors overlap, adding their areas counts the shared region twice. Sketch the shape and decide what to add and what to subtract before writing any formula.
A buoy B is 8 km from a port P on a bearing of 040°. A fishing boat F is on a bearing of 110° from P and is 7.8 km from the buoy. All points are at sea level.
Figure: the two possible positions F₁ and F₂ of the boat (not to scale)
(a)
Write down the size of angle BPF.
(1)
(b)
Show that there are two possible sizes of angle PFB, and find them to 1 decimal place.
(4)
(c)
Find the two possible distances of the boat from the port.
(4)
(d)
The boat is in fact the further of the two possible distances from the port. Find the bearing of the buoy from the boat.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
110° − 40° = 70°
B1
Part (b)
sin F/8 = sin 70°/7.8
M1
sin F = 0.96379…
A1
F = 74.5°
A1
or F = 180° − 74.5° = 105.5°, valid because 70° + 105.5° < 180°
Common error: Forgetting the second triangle. The side opposite the 70° angle (7.8 km) is shorter than the other given side (8 km), so the sine rule can give an obtuse answer as well.
The curve C1 has equation y = sin(x + c), where c is a constant with 0 < c < π. For 0 ≤ x ≤ 2π, the only maximum point of C1 has x-coordinate π/3. The curve C2 has equation y = sin 2x, 0 ≤ x ≤ 2π.
(a)
Find the value of c.
(2)
(b)
Find the exact coordinates of the points where C1 meets the coordinate axes for 0 ≤ x ≤ 2π.
(3)
(c)
State the period of C2 and the coordinates of its maximum points.
(2)
(d)
Show that x = π/6 is a solution of the equation sin 2x = sin(x + c).
(2)
(e)
By considering the graphs of C1 and C2 on the same axes, state the number of solutions of sin 2x = sin(x + c) for 0 ≤ x ≤ 2π.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
π/3 + c = π/2
M1
c = π/6
A1
Part (b)
(0, ½)
B1
(5π/6, 0)
B1
(11π/6, 0)
B1
Part (c)
period π
B1
(π/4, 1) and (5π/4, 1)
B1
Part (d)
LHS = sin(π/3); RHS = sin(π/6 + π/6) = sin(π/3)
M1
both equal √3/2, so x = π/6 is a solution
A1
Part (e)
4
B1
Model sketch
Figure: model sketch of C₁ (c = π/6) and C₂: four points of intersection for 0 ≤ x ≤ 2π
Exam tip: On a sketch with two trig curves, mark where each one crosses the axes and reaches its maxima. Counting crossings is then reliable, especially near the ends of the interval.
Figure 1 shows a plot of land in the shape of a quadrilateral ABCD. The lengths of its four sides and the size of angle ABC are marked on the figure. The diagonal AC is shown dashed.
(a)
Find the length of the diagonal AC.
(3)
(b)
Find the size of angle ADC, giving your answer to 1 decimal place.
(3)
(c)
Find the area of the plot.
(3)
(d)
Find the size of angle BAD, giving your answer to 1 decimal place.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
AC2 = 7.52 + 9.02 − 2 × 7.5 × 9.0 × cos 108°
M1
AC2 = 178.97
A1
AC = 13.4 m (13.378)
A1
Part (b)
cos D = (6.02 + 8.42 − 178.97)/(2 × 6.0 × 8.4)
M1
cos D = −0.7183
A1
angle ADC = 135.9°
A1
Part (c)
triangle ABC: ½ × 7.5 × 9.0 × sin 108° = 32.10
M1
triangle ACD: ½ × 6.0 × 8.4 × sin 135.9° = 17.53
M1
total = 49.6 m2 (49.63)
A1
Part (d)
sin(angle BAC) = 9.0 sin 108°/13.378, angle BAC = 39.8° (acute, as the obtuse angle of triangle ABC is at B)
M1
sin(angle CAD) = 6.0 sin 135.9°/13.378, angle CAD = 18.2°
M1
angle BAD = 39.78° + 18.18° = 58.0°
A1
Exam tip: Keep AC2 = 178.97… in the calculator and use it in the second triangle. Rounding AC to 13.4 first changes angle ADC by about half a degree.
Two circular spotlight beams light a flat stage floor. Figure 1 shows the two circles of light: one has centre P and radius 5 m, the other has centre Q and radius 4 m, and PQ = 6 m. The circles meet at X and Y. The shaded region, lit by both beams, is bounded by the two arcs XY.
(a)
Show that angle XPQ = 0.723 radians, to 3 significant figures.
(2)
(b)
Find the size of angle XQY, in radians.
(3)
(c)
Find the length of the chord XY.
(2)
(d)
Find the area of the region lit by both beams.
(4)
(e)
Find the perimeter of the region lit by both beams.
XY = 2 × 5 sin 0.72273 (or the cosine rule in triangle PXY)
M1
= 6.61 m (6.614)
A1
Part (d)
segment cut off circle P: ½ × 52 × (1.4455 − sin 1.4455)
M1
= 5.666
A1
segment cut off circle Q: ½ × 42 × (1.9468 − sin 1.9468) = 8.133
M1
total = 13.8 m2 (13.80)
A1
Part (e)
5 × 1.4455 + 4 × 1.9468
M1
= 15.0 m (15.01)
A1
Exam tip: The overlap of two circles is two segments, one from each circle, on opposite sides of the common chord XY. Each segment uses its own radius and its own angle at the centre: ½r2(θ − sin θ).
Figure 1 shows a radar station R, which detects ships that are within 18 km of R. The point P is 25 km from R on a bearing of 300°. A ship sails from P due east, on a bearing of 090°, at a constant speed of 15 km h−1. It comes within range of the radar at E and goes out of range at L. Treat the ship as a point moving in a horizontal plane.
(a)
Show that angle RPL = 30°.
(2)
(b)
Show that sin(angle PLR) = 0.6944 to 4 decimal places.
(2)
(c)
Find the distances PE and PL.
(4)
(d)
Find the length of time for which the ship is within range of the radar, giving your answer to the nearest minute.
(2)
(e)
Find the bearing of L from R.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
bearing of R from P = 300° − 180° = 120°
M1
angle RPL = 120° − 90° = 30°
A1
Part (b)
sin(angle PLR)/25 = sin 30°/18
M1
sin(angle PLR) = 12.5/18 = 0.69444…
A1
Part (c)
angle PLR = 43.98° and angle PER = 180° − 43.98° = 136.02°
M1
angle PRL = 106.02°, angle PRE = 13.98°, and distance = 18 sin(angle at R)/sin 30°
M1
PE = 8.70 km (8.699)
A1
PL = 34.6 km (34.60)
A1
(or the cosine rule: x2 − 25√3 x + 301 = 0 with both roots)
—
Part (d)
(34.60 − 8.70)/15 = 1.727 h
M1
= 1 h 44 min (103.6 min)
A1
Part (e)
the bearing of P from R is 300°, and L is clockwise from P through angle PRL
M1
300° + 106.02° = 406.02°
M1
bearing = 406.02° − 360° = 046.0°
A1
Common error: Finding only one point. Both E and L are 18 km from R, so the sine rule gives one acute and one obtuse angle: the obtuse one (136.0°) belongs to the nearer point E.
Figure 1 shows the curve C with equation y = a cos(x − b), 0 ≤ x ≤ 2π, where a and b are constants with a > 0 and 0 < b < π. The curve meets the y-axis at (0, −1.5), and its only maximum point in the interval is (2π/3, 3).
(a)
Write down the value of a and find the value of b. Verify that your values give the point (0, −1.5).
(3)
(b)
Find the exact coordinates of the points where C meets the x-axis, and of the minimum point of C.
(3)
(c)
The curve C can be obtained from the graph of y = cos x by two transformations. Describe each one fully.
(4)
(d)
Find the values of k for which the equation a cos(x − b) = k has exactly one solution for 0 ≤ x ≤ 2π.
(2)
(e)
State the number of solutions of the equation a cos(x − b) = −1.5 for 0 ≤ x ≤ 2π.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
a = 3
B1
the maximum is where x − b = 0, so b = 2π/3
B1
3 cos(−2π/3) = 3 × (−½) = −1.5 ✓
B1
Part (b)
x − 2π/3 = −π/2 or π/2: (π/6, 0)
B1
(7π/6, 0)
B1
minimum (5π/3, −3)
B1
Part (c)
stretch parallel to the y-axis
B1
scale factor 3
B1
translation
B1
by 2π/3 in the positive x-direction, i.e. by the vector (2π/3, 0)
B1
Part (d)
k = 3
B1
k = −3
B1
Part (e)
3 (at x = 0, 4π/3 and 2π)
B1
Exam tip: When counting solutions of f(x) = k on a closed interval, check the end points: here y = −1.5 at both x = 0 and x = 2π, so the line y = −1.5 meets the curve three times, not once.
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