A-Level Triangle trigonometry Practice Questions

Free A-Level Triangle trigonometry practice questions with full step-by-step worked solutions. Covers cosine rule, SAS, sine rule, SSS. Practise exam-style problems and check your method.

cosine ruleSASsine ruleSSSobtuse anglearea of a triangle
A-Level70 questionsStep-by-step solutions
Question 1
3 markseasy
In triangle ABCABC, b=7 cmb = 7\text{ cm}, c=9 cmc = 9\text{ cm} and angle A=40∘A = 40^\circ. Find the length of side aa, giving your answer to 3 significant figures.
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Worked solution

  1. Write down the cosine rule for a side

    a2=b2+c2−2 bccos⁡Aa^2 = b^2 + c^2 - 2\,bc\cos A

    We know two sides and the angle trapped between them, so the cosine rule is the right tool. The side we want, a, sits opposite the known angle A.

  2. Substitute the values you know

    a2=72+92−2×7×9×cos⁡40∘a^2 = 7^2 + 9^2 - 2\times 7\times 9\times\cos 40^{\circ}

    Swap each letter for its number. The two known sides go in the squared terms and the angle between them goes inside the cosine.

  3. Work out the squares and the product

    a2=49+81−126cos⁡40∘a^2 = 49 + 81 - 126\cos 40^{\circ}

    Square each length, and multiply 2, b and c together separately. Breaking it into pieces makes calculator slips less likely.

  4. Evaluate the right-hand side

    a2=33.48a^2 = 33.48

    Make sure the calculator is in degree mode, find the cosine, then combine everything into a single number. This is a squared, not a yet.

  5. Square-root to find the side

    a=33.48=5.79a = \sqrt{33.48} = 5.79

    Because we found a squared, the last move is a square root. A length is always positive, so we keep only the positive value.

  6. State the final answer

    a≈5.79 cma \approx 5.79\text{ cm}

    Rounding to 3 significant figures, side a is about 5.79 cm. It sits opposite the 40° angle, so it should be shorter than the longest side.

Answer
5.79 cm5.79\text{ cm}
Question 2
2 markseasy
In a triangle you are told two sides and the angle between those two sides, and asked to find the third side. Which rule should you use first?
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Worked solution

  1. Look at what the question gives you

    Given: b, c, and the angle A between them\text{Given: } b,\ c,\ \text{and the angle } A \text{ between them}

    First decide what information you have. Here we know two sides and the angle squeezed between them (this is called SAS).

  2. Match the information to a rule

    SAS ⇒ cosine rule\text{SAS}\ \Rightarrow\ \text{cosine rule}

    The sine rule needs a matching side-and-opposite-angle pair, which we do not have. With two sides and the angle between them we must use the cosine rule.

  3. Write the correct rule

    a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc\cos A

    This is the cosine rule for finding the side opposite the known angle. It is the right first step for this triangle.

Answer
Cosine rule\text{Cosine rule}
Question 3
3 marksintermediate
In triangle ABCABC, a=8 cma = 8\text{ cm}, b=10 cmb = 10\text{ cm} and angle A=40∘A = 40^\circ. How many distinct triangles can be drawn with this information, and why?
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Worked solution

  1. Identify the type of information given

    Given: a=8, b=10, A=40∘ (SSA)\text{Given: } a=8,\ b=10,\ A=40^{\circ}\ (\text{SSA})

    We are told two sides and an angle that is NOT between them. This SSA situation is the one that can produce two different triangles.

  2. Recall how the second side can swing

    side b pivots at C; side a swings to meet the base\text{side } b \text{ pivots at } C;\ \text{side } a \text{ swings to meet the base}

    Picture side b fixed at the given angle. Side a (opposite the given angle) can sometimes reach the base line in two places, giving two triangles.

  3. Work out the critical height b·sin A

    bsin⁡A=10sin⁡40∘=6.428b\sin A = 10\sin 40^{\circ} = 6.428

    The shortest possible length for side a to still reach the base is b sin A. Here that critical height is about 6.428.

  4. Compare a with the critical height

    a=8>bsin⁡A=6.428a = 8 > b\sin A = 6.428

    Because a is longer than the critical height, side a definitely reaches the base, so at least one triangle exists.

  5. Compare a with side b

    a=8<b=10a = 8 < b = 10

    Because a is also shorter than b, side a can reach the base on BOTH sides of the foot of the perpendicular.

  6. Apply the rule for two triangles

    bsin⁡A<a<b ⇒ two trianglesb\sin A < a < b\ \Rightarrow\ \text{two triangles}

    When the opposite side a lies between b sin A and b, exactly two different triangles fit the data.

  7. State the conclusion

    6.428<8<10 ⇒ two triangles6.428 < 8 < 10\ \Rightarrow\ \text{two triangles}

    Because a lies between b sin A and b, there are two possible triangles: this is the ambiguous case of the sine rule.

Answer
Two triangles\text{Two triangles}
Question 4
9 markshard
A drone flies from AA on a bearing of 310∘310^\circ for 15 km15\text{ km} to BB, then on a bearing of 050∘050^\circ for 20 km20\text{ km} to CC. Find the bearing of CC from AA, to the nearest 0.1∘0.1^\circ.
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Worked solution

  1. Careful sketch

    A→B:310∘,15; B→C:050∘,20A\to B: 310^{\circ},15;\ B\to C: 050^{\circ},20

    With bearings that cross the north direction, a careful sketch with north lines is essential to get the interior angle right.

  2. Interior angle at B

    ∠ABC=80∘\angle ABC = 80^{\circ}

    Work out the angle inside the triangle at B from the two bearings and the north line; it is 80°.

  3. Cosine rule for AC

    AC2=152+202−2×15×20cos⁡80∘AC^2 = 15^2 + 20^2 - 2\times 15\times 20\cos 80^{\circ}

    Two legs and the included angle give AC.

  4. Evaluate AC²

    AC2=520.8AC^2 = 520.8

    Combine the terms carefully with the calculator in degree mode.

  5. Distance AC

    AC=520.8=22.8 kmAC = \sqrt{520.8} = 22.8\text{ km}

    Square root the result for the straight-line distance.

  6. Angle at A by the sine rule

    sin⁡A=20sin⁡80∘22.8\sin A = \dfrac{20\sin 80^{\circ}}{22.8}

    Find the triangle's angle at A, needed to fix the bearing of C from A.

  7. Evaluate angle A

    A=59.7∘A = 59.7^{\circ}

    Inverse sine gives the angle at A.

  8. Use the north line at A

    first leg points along 310∘\text{first leg points along } 310^{\circ}

    Set up the bearing from the north line at A, remembering the first leg AB is on bearing 310°.

  9. Combine with the first bearing

    bearing of C from A=9.7∘\text{bearing of } C \text{ from } A = 9.7^{\circ}

    Combining the angle at A with the direction of the first leg (with a clear diagram) gives the final bearing 9.7°.

  10. State the final answer

    Bearing of C from A≈9.7∘\text{Bearing of } C \text{ from } A \approx 9.7^{\circ}

    The bearing of C from A is about 9.7°.

Answer
9.7∘9.7^{\circ}
Question 5
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}

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