IB Mathematics SL 3.3 Applications of right and non-right angled trigonometry AI SL Paper 1- Exam Style Questions- New Syllabus
Question
This question uses trigonometry to analyze distances and speed in a real-world scenario. A helicopter’s position and a swimmer’s observations at different angles are used to calculate distances and velocity.
The diagram below shows a helicopter hovering at point H, 380 m vertically above a lake. Point A is the point on the surface of the lake, directly below the helicopter (not to scale).
Alex is swimming at a constant speed in the direction of point A. Alex observes the helicopter from point C as they look upward at an angle of 25°. After 15 minutes, Alex is at point B and observes the same helicopter at an angle of 40°.
(a) State the size of the angle of depression from H to C. [1]
(b) Determine the distance from A to C. [2]
(c) Determine the distance from B to C. [2]
(d) Determine Alex’s speed, in metres per hour. [2]
▶️ Answer/Explanation
Markscheme
(a)
Angle of elevation from C to H is 25°. By alternate interior angles, the angle of depression from H to C is:
\[ 25^\circ \]
Answer: \( 25^\circ \) A1
[1 mark]
Angle of elevation from C to H is 25°. By alternate interior angles, the angle of depression from H to C is:
\[ 25^\circ \]
Answer: \( 25^\circ \) A1
[1 mark]
(b)
In \( \triangle HAC \), \( HA = 380 \, \text{m} \), angle at C = 25°. Use:
\[ \begin{aligned} \tan 25^\circ &= \frac{380}{AC} \\ AC &= \frac{380}{\tan 25^\circ} \approx \frac{380}{0.466307} \approx 814.912 \, \text{m} \approx 815 \, \text{m} \end{aligned} \]
Answer: \( 815 \, \text{m} \) M1 A1
[2 marks]
In \( \triangle HAC \), \( HA = 380 \, \text{m} \), angle at C = 25°. Use:
\[ \begin{aligned} \tan 25^\circ &= \frac{380}{AC} \\ AC &= \frac{380}{\tan 25^\circ} \approx \frac{380}{0.466307} \approx 814.912 \, \text{m} \approx 815 \, \text{m} \end{aligned} \]
Answer: \( 815 \, \text{m} \) M1 A1
[2 marks]
(c)
Method 1:
In \( \triangle HAB \), \( HA = 380 \, \text{m} \), angle at B = 40°. Use:
\[ \begin{aligned} \tan 40^\circ &= \frac{380}{AB} \\ AB &= \frac{380}{\tan 40^\circ} \approx \frac{380}{0.839099} \approx 452.866 \, \text{m} \end{aligned} \]
Since points A, B, C are collinear:
\[ BC = AC – AB \approx 814.912 – 452.866 \approx 362.046 \, \text{m} \approx 362 \, \text{m} \]
Method 2:
In \( \triangle HBC \), angle at H = \( 40^\circ – 25^\circ = 15^\circ \).
Compute \( HB \): \( HB = \frac{380}{\sin 40^\circ} \approx \frac{380}{0.642787} \approx 591.175 \, \text{m} \).
Use sine rule:
\[ \begin{aligned} \frac{BC}{\sin 15^\circ} &= \frac{591.175}{\sin 25^\circ} \\ BC &\approx \frac{591.175 \times \sin 15^\circ}{\sin 25^\circ} \approx \frac{591.175 \times 0.258819}{0.422618} \approx 362.046 \, \text{m} \approx 362 \, \text{m} \end{aligned} \]
Answer: \( 362 \, \text{m} \) M1 A1
[2 marks]
Method 1:
In \( \triangle HAB \), \( HA = 380 \, \text{m} \), angle at B = 40°. Use:
\[ \begin{aligned} \tan 40^\circ &= \frac{380}{AB} \\ AB &= \frac{380}{\tan 40^\circ} \approx \frac{380}{0.839099} \approx 452.866 \, \text{m} \end{aligned} \]
Since points A, B, C are collinear:
\[ BC = AC – AB \approx 814.912 – 452.866 \approx 362.046 \, \text{m} \approx 362 \, \text{m} \]
Method 2:
In \( \triangle HBC \), angle at H = \( 40^\circ – 25^\circ = 15^\circ \).
Compute \( HB \): \( HB = \frac{380}{\sin 40^\circ} \approx \frac{380}{0.642787} \approx 591.175 \, \text{m} \).
Use sine rule:
\[ \begin{aligned} \frac{BC}{\sin 15^\circ} &= \frac{591.175}{\sin 25^\circ} \\ BC &\approx \frac{591.175 \times \sin 15^\circ}{\sin 25^\circ} \approx \frac{591.175 \times 0.258819}{0.422618} \approx 362.046 \, \text{m} \approx 362 \, \text{m} \end{aligned} \]
Answer: \( 362 \, \text{m} \) M1 A1
[2 marks]
(d)
Distance \( BC \approx 362.046 \, \text{m} \), time = 15 minutes = \( \frac{15}{60} = 0.25 \, \text{hours} \).
Speed:
\[ \begin{aligned} \text{Speed} &= \frac{\text{distance}}{\text{time}} = \frac{362.046}{0.25} \approx 1448.184 \, \text{m/h} \approx 1450 \, \text{m/h} \end{aligned} \]
Answer: \( 1450 \, \text{m/h} \) M1 A1
[2 marks]
Distance \( BC \approx 362.046 \, \text{m} \), time = 15 minutes = \( \frac{15}{60} = 0.25 \, \text{hours} \).
Speed:
\[ \begin{aligned} \text{Speed} &= \frac{\text{distance}}{\text{time}} = \frac{362.046}{0.25} \approx 1448.184 \, \text{m/h} \approx 1450 \, \text{m/h} \end{aligned} \]
Answer: \( 1450 \, \text{m/h} \) M1 A1
[2 marks]
Total Marks: 7