Hard A-Level Triangle trigonometry Questions

Challenging, exam-style A-Level Triangle trigonometry questions with worked solutions. Stretch yourself on the hardest ambiguous case, sine rule, cosine rule, area of a triangle problems.

ambiguous casesine rulecosine rulearea of a trianglebearingscomponents
A-Level34 questionsStep-by-step solutions
Question 1
12 markschallenging
A survey drone flies from AA: 20 km20\text{ km} on bearing 060∘060^\circ to BB, 15 km15\text{ km} on bearing 140∘140^\circ to CC, then 18 km18\text{ km} on bearing 250∘250^\circ to DD. Find the straight-line distance ADAD, to 3 significant figures.
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Worked solution

  1. Resolve every leg

    60∘,20; 140∘,15; 250∘,18 (km)60^{\circ},20;\ 140^{\circ},15;\ 250^{\circ},18\ (\text{km})

    For a three-leg journey we resolve every leg into east and north components, add them, and then use Pythagoras and inverse tan for distance and bearing home.

  2. Resolve leg 1

    E:20sin⁡60∘=17.32,N:20cos⁡60∘=10E: 20\sin 60^{\circ} = 17.32,\quad N: 20\cos 60^{\circ} = 10

    East uses sine of the bearing, north uses cosine. Legs heading south or west give negative parts, which we keep.

  3. Resolve leg 2

    E:15sin⁡140∘=9.642,N:15cos⁡140∘=−11.49E: 15\sin 140^{\circ} = 9.642,\quad N: 15\cos 140^{\circ} = -11.49

    East uses sine of the bearing, north uses cosine. Legs heading south or west give negative parts, which we keep.

  4. Resolve leg 3

    E:18sin⁡250∘=−16.91,N:18cos⁡250∘=−6.156E: 18\sin 250^{\circ} = -16.91,\quad N: 18\cos 250^{\circ} = -6.156

    East uses sine of the bearing, north uses cosine. Legs heading south or west give negative parts, which we keep.

  5. Sum the east parts

    Etotal=17.32+9.642+(−16.91)E_{\text{total}} = 17.32 + 9.642 + (-16.91)

    Line up the three east components, keeping the negative one from the westward leg.

  6. Total east displacement

    Etotal=10.05 kmE_{\text{total}} = 10.05\text{ km}

    The drone finishes slightly east of A.

  7. Sum the north parts

    Ntotal=10+(−11.49)+(−6.156)N_{\text{total}} = 10 + (-11.49) + (-6.156)

    Do the same for north, where the last two legs point south (negative).

  8. Total north displacement

    Ntotal=−7.647 kmN_{\text{total}} = -7.647\text{ km}

    The total north displacement is negative, so the drone ends up south of A.

  9. Distance from start

    AD=10.052+−7.6472AD = \sqrt{10.05^2 + -7.647^2}

    Pythagoras on the total east and north displacement gives the straight-line distance AD.

  10. Evaluate the distance

    AD=12.6 kmAD = 12.6\text{ km}

    Work out the square root for the straight-line distance from A to D.

  11. Reference angle from north

    θ=tan⁡−1 ⁣(∣Etotal∣∣Ntotal∣)=52.7∘\theta = \tan^{-1}\!\left(\dfrac{|E_{\text{total}}|}{|N_{\text{total}}|}\right) = 52.7^{\circ}

    The acute angle the line AD makes with the north–south direction comes from inverse tan of east over north.

  12. Pick the correct quadrant

    E>0, N<0⇒south-eastE>0,\ N<0 \Rightarrow \text{south-east}

    A positive east and negative north place D to the south-east of A, so the bearing is between 90° and 180°.

  13. Bearing of D from A

    bearing of D=180∘−52.7∘=127.3∘\text{bearing of } D = 180^{\circ} - 52.7^{\circ} = 127.3^{\circ}

    In the south-east quadrant the bearing is 180° minus the reference angle.

  14. Bearing to return home

    bearing of A from D=307.3∘\text{bearing of } A \text{ from } D = 307.3^{\circ}

    To fly straight back, reverse the direction; the bearing of the start from the finish is about 307.3°.

  15. Compare with the distance flown

    20+15+18=53 km>12.6 km20 + 15 + 18 = 53\text{ km} > 12.6\text{ km}

    The total path length (53 km) is far greater than the direct distance home, as expected for a winding route.

  16. State the final answer

    AD≈12.6 km, home bearing≈307.3∘AD \approx 12.6\text{ km},\ \text{home bearing} \approx 307.3^{\circ}

    The finish is about 12.6 km from the start, on a return bearing of about 307.3°.

Answer
12.6 km12.6\text{ km}
Question 2
12 markschallenging
A triangular plot ABCABC has angle A=40∘A = 40^\circ, angle B=75∘B = 75^\circ and side a=50 ma = 50\text{ m}. Find the other two sides, the area of the plot, and the total length of fencing needed to enclose it. Give the perimeter to 3 significant figures.
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Worked solution

  1. Write down what you know

    A=40∘, B=75∘, a=50 mA=40^{\circ},\ B=75^{\circ},\ a=50\text{ m}

    A surveyor knows two angles and one side of a triangular plot. From this we can find every side, the area, and the length of fencing needed (the perimeter).

  2. C = 180^{\circ} - 40^{\circ} - 75^{\circ}

    C=180∘−40∘−75∘=65∘C = 180^{\circ} - 40^{\circ} - 75^{\circ} = 65^{\circ}

    The three angles inside any triangle always add up to 180°, so we subtract the two known angles to find the third.

  3. Write the sine rule with sides on top

    asin⁡A=bsin⁡B\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

    The sine rule links each side to the sine of the angle opposite it. We put the sides on top because we are hunting for a side.

  4. Substitute the known side and angles

    50sin⁡40∘=bsin⁡75∘\dfrac{50}{\sin 40^{\circ}} = \dfrac{b}{\sin 75^{\circ}}

    Fill in the side you know with its opposite angle, and pair the unknown side with its opposite angle.

  5. Rearrange to make the unknown side the subject

    b=50sin⁡75∘sin⁡40∘b = \dfrac{50\sin 75^{\circ}}{\sin 40^{\circ}}

    Multiply both sides by the sine of the angle opposite the unknown. This leaves the side we want on its own.

  6. Work it out

    b=75.1b = 75.1

    Type it straight into the calculator (in degree mode). This is the length of the side.

  7. Write the sine rule with sides on top

    asin⁡A=csin⁡C\dfrac{a}{\sin A} = \dfrac{c}{\sin C}

    The sine rule links each side to the sine of the angle opposite it. We put the sides on top because we are hunting for a side.

  8. Substitute the known side and angles

    50sin⁡40∘=csin⁡65∘\dfrac{50}{\sin 40^{\circ}} = \dfrac{c}{\sin 65^{\circ}}

    Fill in the side you know with its opposite angle, and pair the unknown side with its opposite angle.

  9. Rearrange to make the unknown side the subject

    c=50sin⁡65∘sin⁡40∘c = \dfrac{50\sin 65^{\circ}}{\sin 40^{\circ}}

    Multiply both sides by the sine of the angle opposite the unknown. This leaves the side we want on its own.

  10. Work it out

    c=70.5c = 70.5

    Type it straight into the calculator (in degree mode). This is the length of the side.

  11. Write the area formula

    Area=12 absin⁡C\text{Area} = \tfrac{1}{2}\,ab\sin C

    When we know two sides and the angle squeezed between them, this formula gives the area straight away. C is the angle between sides a and b.

  12. Substitute the two sides and the included angle

    Area=12×50×75.1357×sin⁡65∘\text{Area} = \tfrac{1}{2}\times 50\times 75.1357\times\sin 65^{\circ}

    Replace the letters with the numbers. Only the angle that lies between the two sides may be used here.

  13. State the area

    Area=1700 cm2\text{Area} = 1700\text{ cm}^2

    This is the area. The units are square centimetres because we multiplied two lengths.

  14. Add up the perimeter

    P=50+75.1+70.5=196P = 50 + 75.1 + 70.5 = 196

    The perimeter (total fencing) is the sum of the three sides.

  15. State the final answer

    b≈75.1, c≈70.5, Area≈1700 m2, P≈196 mb \approx 75.1,\ c \approx 70.5,\ \text{Area} \approx 1700\text{ m}^2,\ P \approx 196\text{ m}

    The plot is fully solved: the area is about 1700 m² and it needs about 196 m of fencing.

Answer
196 m196\text{ m}
Question 3
12 markschallenging
A triangle has angle A=48∘A = 48^\circ, side b=15 cmb = 15\text{ cm} and area 70 cm270\text{ cm}^2. Find the third side cc, side aa, the other two angles, the perimeter, and the radii of the inscribed and circumscribed circles. Give the perimeter to 3 significant figures.
Show worked solution

Worked solution

  1. Write down what you know

    A=48∘, b=15, Area=70A=48^{\circ},\ b=15,\ \text{Area}=70

    We work backwards: the area gives the missing side c, the cosine rule gives a, then we find the remaining angles and the inscribed and circumscribed circle radii.

  2. Use the area to find c

    70=12×15×c×sin⁡48∘70 = \tfrac{1}{2}\times 15\times c\times\sin 48^{\circ}

    Two sides and the included angle appear in the area formula; only c is unknown.

  3. Rearrange for c

    c=2×7015sin⁡48∘=12.6 cmc = \dfrac{2\times 70}{15\sin 48^{\circ}} = 12.6\text{ cm}

    Make c the subject and evaluate.

  4. Cosine rule for a

    a2=152+12.62−2×15×12.6cos⁡48∘a^2 = 15^2 + 12.6^2 - 2\times 15\times 12.6\cos 48^{\circ}

    Now we know two sides (b and c) and the angle A between them, so the cosine rule gives a.

  5. Evaluate a²

    a2=130.6a^2 = 130.6

    Work out the right-hand side.

  6. Square root

    a=11.4 cma = 11.4\text{ cm}

    Take the square root for side a.

  7. Sine rule for angle B

    sin⁡B=15sin⁡48∘11.4=0.9753\sin B = \dfrac{15\sin 48^{\circ}}{11.4} = 0.9753

    With all three sides and angle A known, the sine rule gives angle B.

  8. Evaluate angle B

    B=77.3∘B = 77.3^{\circ}

    Take the inverse sine; b is the longest side so B is the largest angle, but still acute here.

  9. Third angle

    C=180∘−48∘−77.3∘=54.7∘C = 180^{\circ} - 48^{\circ} - 77.3^{\circ} = 54.7^{\circ}

    The angle sum gives the last angle C.

  10. Check the area a different way

    12×11.4×15×sin⁡54.7∘=70 cm2\tfrac{1}{2}\times 11.4\times 15\times\sin 54.7^{\circ} = 70\text{ cm}^2

    Using a different pair of sides and their included angle returns 70 cm², confirming the sides.

  11. Add up the perimeter

    P=11.4+15+12.6=39 cmP = 11.4 + 15 + 12.6 = 39\text{ cm}

    The perimeter is the sum of all three sides.

  12. Semi-perimeter

    s=392=19.5 cms = \dfrac{39}{2} = 19.5\text{ cm}

    Half the perimeter is needed for the inradius.

  13. Inradius

    r=Areas=7019.5=3.59 cmr = \dfrac{\text{Area}}{s} = \dfrac{70}{19.5} = 3.59\text{ cm}

    The inradius is the area divided by the semi-perimeter.

  14. Circumradius

    R=a2sin⁡A=11.42sin⁡48∘=7.69 cmR = \dfrac{a}{2\sin A} = \dfrac{11.4}{2\sin 48^{\circ}} = 7.69\text{ cm}

    The extended sine rule gives the radius of the circle through all three vertices.

  15. State the final answer

    P≈39 cmP \approx 39\text{ cm}

    The perimeter of the triangle is about 39 cm (with r ≈ 3.59 cm and R ≈ 7.69 cm).

Answer
39 cm39\text{ cm}
Question 4
12 markschallenging
Standing at AA, the angle of elevation of a hilltop TT is 20∘20^\circ. After walking 100 m100\text{ m} on level ground directly towards the hill to BB, the angle of elevation is 35∘35^\circ. Find the height of the hilltop above the ground, to 3 significant figures.
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Worked solution

  1. Set up the vertical triangle

    walk 100 m, elevations 20∘,35∘\text{walk } 100\text{ m},\ \text{elevations } 20^{\circ},35^{\circ}

    Two elevation angles from points 100 m apart create a triangle in a vertical plane. We solve the triangle ABT with the sine rule, then drop a vertical to get the height.

  2. Angle at A in the triangle

    ∠TAB=20∘\angle TAB = 20^{\circ}

    At the first point A the line of sight to the summit makes 20° with the horizontal AB.

  3. Angle at B in the triangle

    ∠TBA=180∘−35∘=145∘\angle TBA = 180^{\circ} - 35^{\circ} = 145^{\circ}

    At B the elevation is 35° above the horizontal, but inside triangle ABT the angle is measured back toward A, giving 145°.

  4. Angle at the summit

    ∠ATB=180∘−20∘−145∘=15∘\angle ATB = 180^{\circ} - 20^{\circ} - 145^{\circ} = 15^{\circ}

    The three angles of the triangle add to 180°; equivalently 35° − 20° = 15°.

  5. Sine rule for BT

    BTsin⁡20∘=100sin⁡15∘\dfrac{BT}{\sin 20^{\circ}} = \dfrac{100}{\sin 15^{\circ}}

    The 100 m walk AB is opposite the 15° summit angle; BT is opposite the 20° angle at A.

  6. Solve for BT

    BT=100sin⁡20∘sin⁡15∘=132 mBT = \dfrac{100\sin 20^{\circ}}{\sin 15^{\circ}} = 132\text{ m}

    Rearrange and evaluate the sine rule to find the slant distance BT.

  7. Sine rule for AT

    AT=100sin⁡145∘sin⁡15∘=222 mAT = \dfrac{100\sin 145^{\circ}}{\sin 15^{\circ}} = 222\text{ m}

    The distance from A to the summit is opposite the 145° angle at B.

  8. Drop a vertical from the summit

    right-angled triangle at the base\text{right-angled triangle at the base}

    Let F be the foot of the summit. Triangle BFT is right-angled at F, with BT as the hypotenuse and the 35° elevation at B.

  9. Height using BT

    h=BTsin⁡35∘=132×sin⁡35∘h = BT\sin 35^{\circ} = 132\times\sin 35^{\circ}

    The vertical height h is opposite the 35° angle in that right-angled triangle.

  10. Evaluate the height

    h=75.8 mh = 75.8\text{ m}

    Multiply to find the height of the summit above the horizontal.

  11. Check using AT

    h=ATsin⁡20∘=222×sin⁡20∘=75.8 mh = AT\sin 20^{\circ} = 222\times\sin 20^{\circ} = 75.8\text{ m}

    Using the line of sight from A gives the same height, a reassuring check.

  12. Horizontal distance from B

    BF=BTcos⁡35∘=108 mBF = BT\cos 35^{\circ} = 108\text{ m}

    The horizontal distance from B to the foot of the summit uses cosine of 35°.

  13. Horizontal distance from A

    AF=ATcos⁡20∘=208 mAF = AT\cos 20^{\circ} = 208\text{ m}

    From A the horizontal distance is longer; note AF − BF ≈ 100 m as expected.

  14. Consistency of the walk

    208−108≈100 m208 - 108 \approx 100\text{ m}

    The difference of the two horizontal distances recovers the 100 m walk, confirming the whole solution.

  15. State the final answer

    h≈75.8 mh \approx 75.8\text{ m}

    The summit is about 75.8 m above the level of the walkers.

Answer
75.8 m75.8\text{ m}
Question 5
12 markschallenging
A triangle has two sides of length x cmx\text{ cm} and (x+2) cm(x+2)\text{ cm} with an angle of 30∘30^\circ between them, and its area is 24 cm224\text{ cm}^2. Form a quadratic equation and solve it to find xx, then find the third side, the other two angles and the perimeter (to 3 significant figures).
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Worked solution

  1. Write down what you know

    Area=24, sides x,x+2, angle 30∘\text{Area}=24,\ \text{sides } x, x+2,\ \text{angle } 30^{\circ}

    The area formula gives an equation in x. With the exact value sin 30° = 1/2 it becomes a neat quadratic; once we have x we solve the whole triangle.

  2. Write the area formula

    24=12 x(x+2)sin⁡30∘24 = \tfrac{1}{2}\,x(x+2)\sin 30^{\circ}

    Two sides are x and x+2 with the 30° angle between them.

  3. Use sin 30° = 1/2

    24=12 x(x+2)×12=14x(x+2)24 = \tfrac{1}{2}\,x(x+2)\times\tfrac{1}{2} = \tfrac{1}{4}x(x+2)

    Because sin 30° = 1/2 exactly, the right-hand side simplifies to a quarter of x(x+2).

  4. Multiply out

    96=x2+2x96 = x^2 + 2x

    Multiply both sides by 4, then expand the bracket.

  5. Form a quadratic

    x2+2x−96=0x^2 + 2x - 96 = 0

    Rearrange so one side is zero, ready to factorise.

  6. Factorise

    (x−8)(x+12)=0(x - 8)(x + 12) = 0

    Find two numbers multiplying to −96 and adding to 2: they are +8 and −12.

  7. Check with the quadratic formula

    x=−2±4+3842=−2±3882x = \dfrac{-2\pm\sqrt{4 + 384}}{2} = \dfrac{-2\pm\sqrt{388}}{2}

    As a check, the discriminant is 2² − 4(1)(−96) = 388, and √388 ≈ 19.70.

  8. Choose the valid root

    x=8(x=−12 rejected, length>0)x = 8 \quad (x = -12 \text{ rejected, length}>0)

    A length cannot be negative, so we discard x = −12.

  9. State the two sides

    x=8 cm, x+2=10 cmx = 8\text{ cm},\ x+2 = 10\text{ cm}

    The two given sides are therefore 8 cm and 10 cm.

  10. Third side by the cosine rule

    y2=82+102−2×8×10cos⁡30∘y^2 = 8^2 + 10^2 - 2\times 8\times 10\cos 30^{\circ}

    The 30° angle is between the two sides, so the cosine rule gives the side opposite it.

  11. Evaluate y²

    y2=164−160cos⁡30∘=25.44y^2 = 164 - 160\cos 30^{\circ} = 25.44

    Work out the right-hand side in degree mode.

  12. Third side

    y=25.44=5.04 cmy = \sqrt{25.44} = 5.04\text{ cm}

    Square root for the length of the third side.

  13. Angle opposite the 8 cm side

    sin⁡θ=8sin⁡30∘5.04, θ=52.5∘\sin\theta = \dfrac{8\sin 30^{\circ}}{5.04},\ \theta = 52.5^{\circ}

    The sine rule gives the angle facing the 8 cm side.

  14. Third angle

    180∘−30∘−52.5∘=97.5∘180^{\circ} - 30^{\circ} - 52.5^{\circ} = 97.5^{\circ}

    The angles sum to 180°; this one is obtuse, facing the longest side.

  15. Perimeter

    P=8+10+5.04=23 cmP = 8 + 10 + 5.04 = 23\text{ cm}

    Add the three sides for the perimeter.

  16. State the final answer

    x=8 cmx = 8\text{ cm}

    The side x is 8 cm (the sides are 8 cm and 10 cm, the third side about 5.04 cm).

Answer
88

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