Sine and cosine rules Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Sine and cosine rules questions. See exactly how to solve problems on sine rule, missing side, rounding to a given accuracy, missing angle.

sine rulemissing siderounding to a given accuracymissing anglecosine ruleobtuse angle
GCSE Higher70 questionsStep-by-step solutions
Question 1
2 markseasy
In triangle ABCABC, angle A=40∘A = 40^\circ, angle B=75∘B = 75^\circ and side a=9a = 9 cm. Use the sine rule to work out the length of side bb. Give your answer correct to 11 decimal place.

Worked solution

  1. State the sine rule.

    asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

    The sine rule links each side of a triangle to the sine of the angle opposite it: asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. Use it whenever you know a side and the angle opposite it.

  2. Pick out the pair of sides and angles that the question needs.

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

    Side bb is opposite angle BB, and side aa is opposite angle AA. Both of those pairings are known or findable, so the sine rule can start.

  3. State the length of side b.

    b=9sin⁡75∘sin⁡40∘=13.5 cmb = \frac{9\sin 75^\circ}{\sin 40^\circ} = 13.5\text{ cm}

    Side bb is 13.513.5 cm, correct to 11 decimal place.

Answer
b=13.5 cmb = 13.5\text{ cm}
Question 2
2 markseasy
In triangle ABCABC, angle A=35∘A = 35^\circ, angle B=80∘B = 80^\circ and side a=12a = 12 cm. Use the sine rule to work out the length of side bb. Give your answer correct to 11 decimal place.

Worked solution

  1. State the sine rule.

    asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

    The sine rule links each side of a triangle to the sine of the angle opposite it: asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. Use it whenever you know a side and the angle opposite it.

  2. Pick out the pair of sides and angles that the question needs.

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

    Side bb is opposite angle BB, and side aa is opposite angle AA. Both of those pairings are known or findable, so the sine rule can start.

  3. State the length of side b.

    b=12sin⁡80∘sin⁡35∘=20.6 cmb = \frac{12\sin 80^\circ}{\sin 35^\circ} = 20.6\text{ cm}

    Side bb is 20.620.6 cm, correct to 11 decimal place.

Answer
b=20.6 cmb = 20.6\text{ cm}
Question 3
2 markseasy
In triangle ABCABC, angle A=50∘A = 50^\circ, angle B=64∘B = 64^\circ and side a=10a = 10 cm. Use the sine rule to work out the length of side bb. Give your answer correct to 11 decimal place.

Worked solution

  1. State the sine rule.

    asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

    The sine rule links each side of a triangle to the sine of the angle opposite it: asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. Use it whenever you know a side and the angle opposite it.

  2. Pick out the pair of sides and angles that the question needs.

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

    Side bb is opposite angle BB, and side aa is opposite angle AA. Both of those pairings are known or findable, so the sine rule can start.

  3. State the length of side b.

    b=10sin⁡64∘sin⁡50∘=11.7 cmb = \frac{10\sin 64^\circ}{\sin 50^\circ} = 11.7\text{ cm}

    Side bb is 11.711.7 cm, correct to 11 decimal place.

Answer
b=11.7 cmb = 11.7\text{ cm}
Question 4
2 markseasy
In triangle ABCABC, angle A=42∘A = 42^\circ, angle B=63∘B = 63^\circ and side a=8a = 8 cm. Use the sine rule to work out the length of side cc. Give your answer correct to 11 decimal place.

Worked solution

  1. State the sine rule.

    asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

    The sine rule links each side of a triangle to the sine of the angle opposite it: asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. Use it whenever you know a side and the angle opposite it.

  2. Pick out the pair of sides and angles that the question needs.

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

    Side cc is opposite angle CC, and side aa is opposite angle AA. Both of those pairings are known or findable, so the sine rule can start.

  3. State the length of side c.

    c=8sin⁡75∘sin⁡42∘=11.5 cmc = \frac{8\sin 75^\circ}{\sin 42^\circ} = 11.5\text{ cm}

    Side cc is 11.511.5 cm, correct to 11 decimal place.

Answer
c=11.5 cmc = 11.5\text{ cm}
Question 5
2 markseasy
In triangle ABCABC, angle A=55∘A = 55^\circ, angle B=65∘B = 65^\circ and side a=14a = 14 cm. Use the sine rule to work out the length of side cc. Give your answer correct to 11 decimal place.

Worked solution

  1. State the sine rule.

    asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

    The sine rule links each side of a triangle to the sine of the angle opposite it: asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. Use it whenever you know a side and the angle opposite it.

  2. Pick out the pair of sides and angles that the question needs.

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

    Side cc is opposite angle CC, and side aa is opposite angle AA. Both of those pairings are known or findable, so the sine rule can start.

  3. State the length of side c.

    c=14sin⁡60∘sin⁡55∘=14.8 cmc = \frac{14\sin 60^\circ}{\sin 55^\circ} = 14.8\text{ cm}

    Side cc is 14.814.8 cm, correct to 11 decimal place.

Answer
c=14.8 cmc = 14.8\text{ cm}

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