Grade 12 Maths · Trigonometry · Show a length equals an expression

Show a length equals an expression Grade 12

Use the sine or cosine rule with letters instead of numbers, then an identity, to show that a length equals a given expression (for example AB = 2x), and then work out its value.

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2D Trigonometry: VariablesKevinmathscience

2D and 3D Trigonometry - Grade 12 November 2022 Maths Exam Paper 2 Memo | Question 7AskPerci

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Question 7.1 · November 2023 · Paper 2 · National (DBE) · 2 marks

In the diagram, S, T and K lie in the same horizontal plane. RS is a vertical tower. The angle of depression from R to K is

β

T \hat{S} K = α

, TS

= p

metres and the area of

Δ ST K

q

^{2}

. $Diagram for this question$

7.1 Determine the length of SK in terms of

p

q

and

α

.(2)

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More past paper questions: show a length equals an expression

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3. Question 7.2.3 · May/June 2024 · Paper 2 · National (DBE) · 3 marks
7.2 In the diagram, points B, C and D lie in the same horizontal plane. $A \hat{D} B = A \hat{C} B = α$ , $C \hat{D} B = θ$ and $D C = k$ units. BD = BC.

Answer the questions that follow. $Diagram for question 7.2$
7.2.3 Determine the area of $Δ BC D$ in terms of $k$ and a single trigonometric ratio of $θ$ .(3)
Show the memo (3 ticks)Memo
1. ✓
  Substitution into the area ruleArea of $Δ BC D = \frac{1}{2} (D C) (B D) (sin C \hat{D} B)$
  $= \frac{1}{2} k (\frac{k}{2 c o s θ}) sin θ$
  1 mark · Use the area rule
2. ✓
  $\frac{s i n θ}{c o s θ} = tan θ$ $\frac{s i n θ}{c o s θ} = tan θ$
  1 mark · Write tan as sin over cos
3. ✓
  Answer $= \frac{1}{4} k^{2} tan θ$
  1 mark · Show a length equals an expression
4. 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)
Show the memo (4 ticks)Memo
1. ✓
  METHOD 1: Using the Sine Rule in the horizontal triangleIn $Δ B A C$ :
  Since $A B = A C$ , $Δ B A C$ is isosceles, so $A \hat{C} B = A \hat{B} C = x$ .
  Sum of angles in a triangle: $B \hat{A} C = 18 0^{\circ} - (x + x) = 18 0^{\circ} - 2 x$ .
  We are given $2 A D = BC$ , and $A D = p$ , so $BC = 2 p$ .
  
  Apply the sine rule:
  $\frac{s i n B A ^ C}{BC} = \frac{s i n A C ^ B}{A B}$
  1 mark · Use the sine rule
2. ✓
  Substitute known values into the Sine Rule $\frac{s i n ( 18 0 ^{\circ} - 2 x )}{2 p} = \frac{s i n x}{\frac{p}{t a n y}}$
  1 mark · Solve a 2D or 3D triangle problem
3. ✓
  Simplify the equation using reduction and double angle identities $\frac{s i n 2 x}{2 p} = sin x \times (\frac{t a n y}{p})$
  $\frac{2 s i n x c o s x}{2 p} = \frac{s i n x t a n y}{p}$
  1 mark · Use a double angle identity
4. ✓
  Solve for cos xMultiply both sides by $p$ and cancel the $2$ s on the LHS:
  $sin x cos x = sin x tan y$
  Divide both sides by $sin x$ (since $sin x \neq = 0$ ):
  $cos x = tan y$
  1 mark · Show a length equals an expression
5. Question 6.3 · September 2026 · Paper 2 · Provincial · 2 marks
6.3 If area $△$ PQR $= \frac{1}{2} pq sin R$ , show that the area $△$ PQR can also be expressed as $\frac{q ^{2} s i n P s i n R}{2 s i n Q}$ .(2)
Show the memo (2 ticks)Memo
1. ✓
  $p = \frac{q s i n P}{s i n Q}$ $\frac{p}{s i n P} = \frac{q}{s i n Q}$
  $p = \frac{q s i n P}{s i n Q}$ ... (2)
  1 mark · Use the sine rule
2. ✓
  SubstitutionSubstitute (2) into area $= \frac{1}{2} pq sin R$ :
  area $= \frac{1}{2} (\frac{q s i n P}{s i n Q}) q sin R = \frac{q ^{2} s i n P s i n R}{2 s i n Q}$
  1 mark · Show a length equals an expression
6. Question 7.1 · September 2024 · Paper 2 · Provincial · 2 marks
In the diagram, B, C and D lie in the same horizontal plane.
$B D = 2 C D$ .
$C \hat{B} D = x$ , $B \hat{C} D = y$ and $C D = b$ meters.
AB is a vertical tower.
The angle of elevation of A from C is $θ$ . $Diagram for this question$
7.1 Show that $sin y = 2 sin x$ .(2)
Show the memo (2 ticks)Memo
1. ✓
  Substitution in sine rule $\frac{s i n y}{2 b} = \frac{s i n x}{b}$
  1 mark · Use the sine rule
2. ✓
  $sin y = \frac{2 b s i n x}{b}$ OR $b sin y = 2 b sin x$ $sin y = \frac{2 b s i n x}{b}$ OR $b sin y = 2 b sin x$
  $sin y = 2 sin x$
  1 mark · Show a length equals an expression

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