Triangle trigonometry Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Triangle trigonometry questions. See exactly how to solve problems on cosine rule, SAS, sine rule, SSS.

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.

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
3 markseasy
A triangle has two sides of length 12 cm12\text{ cm} and 8 cm8\text{ cm} with an angle of 65∘65^\circ between them. Work out the length of the third side, to 3 significant figures.

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=122+82−2×12×8×cos⁡65∘a^2 = 12^2 + 8^2 - 2\times 12\times 8\times\cos 65^{\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=144+64−192cos⁡65∘a^2 = 144 + 64 - 192\cos 65^{\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=126.9a^2 = 126.9

    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=126.9=11.3a = \sqrt{126.9} = 11.3

    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≈11.3 cma \approx 11.3\text{ cm}

    To 3 significant figures the side is 11.3 cm. Check it is a believable length compared with the other two sides.

Answer
11.3 cm11.3\text{ cm}
Question 3
3 markseasy
In triangle ABCABC, angle A=50∘A = 50^\circ, angle B=70∘B = 70^\circ and side a=10 cma = 10\text{ cm}. Find the length of side bb, to 3 significant figures.

Worked solution

  1. 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.

  2. Substitute the known side and angles

    10sin⁡50∘=bsin⁡70∘\dfrac{10}{\sin 50^{\circ}} = \dfrac{b}{\sin 70^{\circ}}

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

  3. Rearrange to make the unknown side the subject

    b=10sin⁡70∘sin⁡50∘b = \dfrac{10\sin 70^{\circ}}{\sin 50^{\circ}}

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

  4. Work it out

    b=12.3b = 12.3

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

  5. State the final answer

    b≈12.3 cmb \approx 12.3\text{ cm}

    To 3 significant figures side b is 12.3 cm. Because 70° is bigger than 50°, side b should be longer than 10 cm, which it is.

Answer
12.3 cm12.3\text{ cm}
Question 4
3 markseasy
In triangle ABCABC, a=15 cma = 15\text{ cm}, angle A=45∘A = 45^\circ and angle B=60∘B = 60^\circ. Calculate the length of side bb, to 3 significant figures.

Worked solution

  1. 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.

  2. Substitute the known side and angles

    15sin⁡45∘=bsin⁡60∘\dfrac{15}{\sin 45^{\circ}} = \dfrac{b}{\sin 60^{\circ}}

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

  3. Rearrange to make the unknown side the subject

    b=15sin⁡60∘sin⁡45∘b = \dfrac{15\sin 60^{\circ}}{\sin 45^{\circ}}

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

  4. Work it out

    b=18.4b = 18.4

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

  5. State the final answer

    b≈18.4 cmb \approx 18.4\text{ cm}

    The side is 18.4 cm to 3 significant figures. The larger angle is opposite the larger side, which is a handy check.

Answer
18.4 cm18.4\text{ cm}
Question 5
3 markseasy
A triangle has sides a=6 cma = 6\text{ cm}, b=7 cmb = 7\text{ cm} and c=8 cmc = 8\text{ cm}. Find the size of angle AA, giving your answer to 1 decimal place.

Worked solution

  1. Write the cosine rule rearranged for an angle

    cos⁡A=b2+c2−a22 bc\cos A = \dfrac{b^2 + c^2 - a^2}{2\,bc}

    When all three sides are known, this rearranged cosine rule finds an angle. The side on its own on top (a) is the one opposite the angle we want.

  2. Substitute the three side lengths

    cos⁡A=72+82−622×7×8\cos A = \dfrac{7^2 + 8^2 - 6^2}{2\times 7\times 8}

    The two sides that touch the angle go on the bottom; all three sides appear on top. Take care to use the side opposite the angle in the last squared term.

  3. Work out the top and the bottom

    cos⁡A=77112\cos A = \dfrac{77}{112}

    Deal with the numerator and denominator separately, then you are left with a simple division.

  4. Divide to find the cosine of the angle

    cos⁡A=0.6875\cos A = 0.6875

    Dividing gives the value of the cosine. If this comes out negative the angle is obtuse (bigger than 90°).

  5. Use inverse cosine to get the angle

    A=cos⁡−1(0.6875)=46.6∘A = \cos^{-1}(0.6875) = 46.6^{\circ}

    Press shift then cos (inverse cosine) to turn the ratio back into an angle. Keep the calculator in degrees.

  6. State the final answer

    A≈46.6∘A \approx 46.6^{\circ}

    The angle is about 46.6°. It is the smallest angle because it is opposite the shortest side, which fits.

Answer
46.6∘46.6^{\circ}

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