Grade 11 Maths · Trigonometry · Solve problems with several triangles

Solve problems with several triangles Grade 11

Work through a diagram with more than one triangle, using the sine, cosine and area rules one triangle at a time, including angles of elevation and depression and 'show that' expressions.

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Grade 11 - Sine and Cosine Rule Application - 2D Trigonometry | Mlungisi NkosiMlungisi Nkosi Maths & Science

Grade 11 - Sine Rule and Cosine Rule Application | Past Exam Question | Mlungisi NkosiMlungisi Nkosi Maths & Science

2D&3D SHAPES TRIGONOMETRY| GRADE 11| PAST EXAM QUESTION| AREA RULE, SINE AND COSINEBACK2BASICS LEARNING INSTITUTE

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Question 6.2 · November 2025 · Paper 2 · Provincial · 3 marks

A man standing on a horizontal plane at point O observes two helicopters positioned at points P and Q. The lines of sight from point O to each helicopter form angles of elevation

θ

and

β

with the horizontal plane. Additionally, a second man located in the helicopter at point P observes the helicopter at point Q, and his line of sight forms an angle of elevation

α

with the horizontal plane at P, measured along line PQ.

ON

⊥

NR and OP

= x

. $Diagram for this question$

6.2 Show that

OQ = \frac{x . s i n ( α + θ )}{s i n ( β + α )}

(3)

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More questions: solve problems with several triangles

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3. Question 7.2.2 · November 2019 · Paper 2 · National (DBE) · 6 marks
7.2 In the diagram below, PTSR is a parallelogram. Q is the midpoint of PR. QS $= 6$ units, PT $= 3$ units, TS $= 8$ units, QT $= x$ units and $\hat{T}_{1} = θ$ .

Answer the questions that follow. $Diagram for question 7.2$
7.2.2 Hence, determine the length of QT.(6)
Show the memo (6 ticks)Memo
1. ✓
  $\hat{Q}_{1} = θ$ and PQ $= 4$ $\hat{Q}_{1} = θ$ (alt $∠$ s, PR $∥$ TS) and PQ $= 4$
  1 mark · Use angles on parallel lines
2. ✓
  Cosine rule in $△$ QTPIn $△$ QTP:
  $3^{2} = 4^{2} + x^{2} - 2 (4) (x) cos θ$
  1 mark · Use the cosine rule
3. ✓
  $cos θ = \frac{7 + x ^{2}}{8 x}$ $8 x cos θ = 4^{2} + x^{2} - 3^{2}$
  $8 x cos θ = x^{2} + 7$
  $cos θ = \frac{7 + x ^{2}}{8 x}$
  1 mark · Use the cosine rule
4. ✓
  Equating $\frac{7 + x ^{2}}{8 x} = \frac{28 + x ^{2}}{16 x}$
  1 mark · Solve problems with several triangles
5. ✓
  Factors $112 x + 16 x^{3} = 224 x + 8 x^{3}$
  $8 x^{3} - 112 x = 0$
  $8 x (x^{2} - 14) = 0$
  1 mark · Take out a common factor
6. ✓
  Answer $x \neq = 0$ , $x = 14$
  1 mark · Solve a quadratic by factorising
4. Question 5.2 · November 2023 · Paper 2 · Provincial · 4 marks
In the diagram below, PQ is a vertical tower. Q, R and S are points on the same horizontal plane. The angle of elevation from R to P is $θ$ . $Q \hat{R} S = 120°$ and PQ = RS = $x$ . $Diagram for this question$
5.2 Show that QS $= x \frac{1}{t a n ^{2} θ} + \frac{1}{t a n θ} + 1$ (4)
Show the memo (4 ticks)Memo
1. ✓
  Cosine rule $QS^{2} = x^{2} + (\frac{x}{t a n θ})^{2} - 2 x (\frac{x}{t a n θ}) cos 120°$
  1 mark · Solve problems with several triangles
2. ✓
  Value of cos 120° $QS^{2} = x^{2} + \frac{x ^{2}}{t a n ^{2} θ} - \frac{2 x ^{2}}{t a n θ} (- \frac{1}{2})$
  1 mark · Use special angle values
3. ✓
  Simplification $QS^{2} = x^{2} + \frac{x ^{2}}{t a n ^{2} θ} + \frac{x ^{2}}{t a n θ}$
  1 mark · Use the cosine rule
4. ✓
  Square rooting $QS = x^{2} (1 + \frac{1}{t a n ^{2} θ} + \frac{1}{t a n θ})$
  $QS = x \frac{1}{t a n ^{2} θ} + \frac{1}{t a n θ} + 1$
  1 mark · Use the cosine rule
5. Practice · NavyBlue · 2 marks
$FT$ is a vertical tower. A and B are two points on level ground in a straight line with F, the foot of the tower. $A B = 40$ m. The angles of elevation of T from A and from B are $2 8^{\circ}$ and $4 1^{\circ}$ respectively.
(a) Show that $A \hat{T} B = 1 3^{\circ}$ .(2)
Show the memo (2 ticks)Memo
1. ✓
  Angle on a straight line $A \hat{B} T = 18 0^{\circ} - 4 1^{\circ} = 13 9^{\circ}$
  1 mark · Solve problems with several triangles
2. ✓
  Angles in $△ A BT$ $A \hat{T} B = 18 0^{\circ} - 2 8^{\circ} - 13 9^{\circ} = 1 3^{\circ}$
  1 mark · Use angle sum of a triangle
6. Practice · Siyavula Grade 11 Maths · Ex 6.14 · 2 marks
$A$ is the highest point of a vertical tower $A T$ . At point $N$ on the tower, $n$ metres from the top of the tower, a bird has made its nest. The angle of inclination from $G$ to point $A$ is $α$ and the angle of inclination from $G$ to point $N$ is $β$ .
(c) Show that the height of the nest from the ground ( $H$ ) can be determined by the formula $H = \frac{n cos α sin β}{sin ( α - β )}$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Sine rule in $△ A GN$ $\frac{GN}{sin ( 90° - α )} = \frac{n}{sin ( α - β )}$ , so $GN = \frac{n cos α}{sin ( α - β )}$
  1 mark · Use the sine rule
2. ✓
  Right-angled $△ GNT$ $H = GN sin β = \frac{n cos α sin β}{sin ( α - β )}$
  1 mark · Solve problems with several triangles
Question adapted from Siyavula Education (CC BY 3.0) · Solution by NavyBlue

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