Grade 12 Maths · Trigonometry · Solve a 2D or 3D triangle problem

Solve a 2D or 3D triangle problem Grade 12

Work through two or more linked triangles, often one flat and one upright, finding the shared side first and then using the sine, cosine or area rule or right-angle trig, for example the height of a tower seen from two points on the ground.

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3D Trigonometry grade 11 and 12Kevinmathscience

Lesson 12: How To Solve 3D Trigonometry Problems | Grade 12 TrigonometryMlungisi Nkosi Maths & Science

How to do 3D Trigonometry - Grade 12 Mathematics (Paper 2)MatheNatic

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Question 8.1 · May/June 2025 · Paper 2 · National (DBE) · 2 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

8.1 Determine AB in terms of p and y.(2)

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

More past paper questions: solve a 2d or 3d triangle problem

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3. Question 7.1.1 · May/June 2023 · Paper 2 · National (DBE) · 2 marks
In the diagram, A, B and C are points in the same horizontal plane. D is a point directly above C, that is DC $⊥$ AC and DC $⊥$ BC. It is given that $A \hat{C} B = 100°$ , $C \hat{A} D = 30°$ , AD $= 20$ units and BC $= 8$ units. $Diagram for this question$
7.1 Calculate the length of:
7.1.1 AC(2)
Show the memo (2 ticks)Memo
1. ✓
  METHOD 1: Trig ratio $\frac{A C}{20} = cos 30°$
  1 mark · Solve a 2D or 3D triangle problem
2. ✓
  METHOD 1: Answer $A C = 20 cos 30°$
  $A C = 103 = 17, 32$ units
  1 mark · Calculate a side using trig
4. Question 8.1 · May/June 2022 · Paper 2 · National (DBE) · 2 marks
FIGURE I shows a ramp leading to the entrance of a building. B, C and D lie on the same horizontal plane. The perpendicular height (AC) of the ramp is $0, 5$ m and the angle of elevation from B to A is $1 5^{\circ}$ . The entrance of the building (AE) is $0, 915$ m wide. $Diagram for this question$
8.1 Calculate the length of AB.(2)
Show the memo (2 ticks)Memo
1. ✓
  Trig ratio $\frac{0 , 5}{A B} = sin 1 5^{\circ}$
  1 mark · Solve a 2D or 3D triangle problem
2. ✓
  Answer $A B = \frac{0 , 5}{s i n 1 5 ^{\circ}} = 1, 93$ m
  1 mark · Calculate a side using trig
5. Question 8.3 · September 2026 · Paper 2 · Provincial · 2 marks
In the diagram, A, B and C are points in the same horizontal plane. AD is a vertical tower anchored at B and C. The angle of elevation of D from B is $25°$ , A $\hat{C}$ B $= 50°$ and B $\hat{D}$ C $= 58°$ .
AC $=$ BC $= 22$ metres.

Calculate the following: $Diagram for this question$
8.3 The length of DB(2)
Show the memo (2 ticks)Memo
1. ✓
  Substitution $cos 25° = \frac{18 , 60}{D B}$
  1 mark · Solve a 2D or 3D triangle problem
2. ✓
  AnswerDB $= \frac{18 , 60}{c o s 25°} = 20, 52$ m
  1 mark · Calculate a side using trig
6. Question 7.4 · September 2025 · Paper 2 · Provincial · 4 marks
In the diagram below, PS is a vertical tower standing on a horizontal plane QRS.

QR $= y$ , $\hat{Q}_{1} = α$ , $\hat{Q}_{2} = θ$ and $Q \hat{S} R = β$ . $Diagram for this question$
7.4 Determine the angle of elevation from R to P.(4)
Show the memo (4 ticks)Memo
1. ✓
  Substitution into sine rule $\frac{s i n α}{RS} = \frac{s i n β}{y}$
  $\frac{s i n 27°}{RS} = \frac{s i n 102°}{116}$
  1 mark · Use the sine rule
2. ✓
  Length of RS $RS = \frac{116 s i n 27°}{s i n 102°} = 53, 84$ m
  1 mark · Use the sine rule
3. ✓
  Correct substitution in ratio of $tan P \hat{R} S$ $tan P \hat{R} S = \frac{PS}{RS} = \frac{141 , 92}{53 , 84}$
  1 mark · Solve a 2D or 3D triangle problem
4. ✓
  Answer $P \hat{R} S = tan^{- 1} (\frac{141 , 92}{53 , 84}) = 69, 22°$
  1 mark · Calculate an angle from a ratio

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