Grade 11 Maths · Trigonometry

Use the sine rule

Use the sine rule a/sin A = b/sin B to find a side or angle when you know a side and the angle opposite it.

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Sine Rule grade 11: IntroductionKevinmathscience

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Sine Rule grade 11: Ambiguous CaseKevinmathscience

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Question 8.2 · May/June 2025 · Paper 2 · National (DBE) · 4 marks

In the diagram, A, B and C lie in the same horizontal plane with

A B = A C

. D is directly above A such that

2 A D = BC

. Also,

A D = p

A \hat{B} C = x

and

D \hat{B} A = y

8.2 Show that

cos x = tan y

.(4)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: use the sine rule

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3. Question 5.3.2 · March 2025 · Provincial · 3 marks
5.3 A piece of land has the form of a quadrilateral ABCD with $A B = 20 m$ , $BC = 12 m$ and $C D = 7 m$ . $\hat{B} = 11 0^{\circ}$ and $\hat{C D A} = 7 1^{\circ}$ . The owner decides to divide the land into two plots by erecting a fence from A to C.
5.3.2 Calculate the size of $D \hat{A} C$ correct to two decimal places.(3)
Show the memo (3 ticks)Memo
1. ✓
  Setup Sine Rule $\frac{s i n D A ^ C}{D C} = \frac{s i n A D ^ C}{A C}$
  1 mark · Use the sine rule
2. ✓
  Substitute values $\frac{s i n D A ^ C}{7} = \frac{s i n 7 1 ^{\circ}}{26 , 61}$
  1 mark · Use the sine rule
3. ✓
  Calculate angle $sin D \hat{A} C = 0, 2487...$
  $D \hat{A} C = 14, 4 0^{\circ}$
  1 mark · Use the sine rule
4. Question 7.2.1 · November 2024 · Paper 2 · Provincial · 2 marks
In the diagram below, circle ACD is given where AC is a diameter. DC is produced to B and AB = 8 units.
$\hat{B} = 30°$ and BC = 5 units. $Diagram for this question$
7.2 Calculate the following:
7.2.1 $\hat{A}_{1}$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Substitution into sine rule $\frac{s i n A ^ _{1}}{5} = \frac{s i n 30°}{4 , 44}$
  1 mark · Use the sine rule
2. ✓
  Answer $sin \hat{A}_{1} = \frac{5 s i n 30°}{4 , 44}$
  $\hat{A}_{1} = 34, 27°$
  1 mark · Use the sine rule
5. Question 4.2 · March 2022 · Provincial · 3 marks
In the diagram below, A, B and C are in the same horizontal plane. P is a point vertically above A. The angle of elevation from B to P is $α$ . $A \hat{C} B = β$ , $A \hat{B} C = θ$ and $BC = 20$ units.
4.2 Prove that $A P = \frac{20 s i n β t a n α}{s i n ( θ + β )}$ (3)
Show the memo (3 ticks)Memo
1. ✓
  Apply Sine Rule in triangle ABC $\frac{A B}{s i n β} = \frac{BC}{s i n ( 18 0 ^{\circ} - ( θ + β ))}$
  1 mark · Use the sine rule
2. ✓
  Express AB $A B = \frac{20 s i n β}{s i n ( θ + β )}$
  1 mark · Use the sine rule
3. ✓
  Substitute into AP formula $A P = (\frac{20 s i n β}{s i n ( θ + β )}) tan α = \frac{20 s i n β t a n α}{s i n ( θ + β )}$
  1 mark · Use the sine rule
6. Question 6.1 · March 2021 · Provincial · 2 marks
In the sketch, $PS$ is the median of $△ PQR$ and thus $QS = SR = x$ . $\hat{Q} = a$ and $Q \hat{P} S = b$ .
6.1 Show that $PS = \frac{x s i n a}{s i n b}$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Apply Sine Rule in Triangle PQS $\frac{PS}{s i n a} = \frac{QS}{s i n b}$
  $\frac{PS}{s i n a} = \frac{x}{s i n b}$
  1 mark · Use the sine rule
2. ✓
  Solve for PS $PS = \frac{x s i n a}{s i n b}$
  1 mark · Use the sine rule

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