Checkpoint 1
Right-angled Trigonometry. Assess your understanding of sine, cosine, tangent, finding unknown sides and angles, angles of elevation and depression, and compass bearings.
Section A - Multiple Choice
2 marks each. Select the best answer.
2 marks In a right-angled triangle, $\sin \theta = \frac{3}{5}$. The hypotenuse is 20 cm. The length of the opposite side is:
2 marks If $\cos \theta = 0.6$, then $\theta$ is approximately:
2 marks A tower casts a shadow 30 m long. The angle of elevation of the sun is $35°$. The height of the tower is closest to:
2 marks The angle of depression from the top of a cliff to a boat is $25°$. If the cliff is 80 m high, the distance from the base of the cliff to the boat is closest to:
2 marks A bearing of $135°$ is equivalent to the compass direction:
2 marks The exact value of $\tan 45°$ is:
2 marks A ship sails $15$ km on a bearing of $070°$. The distance it has travelled east is closest to:
2 marks In a right-angled triangle, the hypotenuse is $17$ cm and one side is $8$ cm. The other side is:
2 marks The back bearing of $210°$ is:
2 marks A ramp makes an angle of $8°$ with the horizontal. If the ramp is $5$ m long, the vertical rise is closest to:
Section B - Short Answer
Show all working. 5 marks each.
Question 11
From the top of a $45$ m tall lighthouse, the angle of depression to a boat is $18°$.
(a) Draw a diagram showing the angle of depression and the right-angled triangle formed. (1 mark)
(b) Calculate the horizontal distance from the base of the lighthouse to the boat. (2 marks)
(c) A second boat is directly behind the first boat, and the angle of depression to the second boat is $12°$. How far apart are the two boats? (2 marks)
Marking Criteria
- 1 mark: Correct diagram with angle of depression shown (equal to angle of elevation from boat)
- 1 mark: Set up trig ratio $\tan 18° = \frac{45}{d}$
- 1 mark: Correct distance $d = \frac{45}{\tan 18°} \approx 138$ m
- 1 mark: Distance to second boat $= \frac{45}{\tan 12°} \approx 212$ m
- 1 mark: Difference $212 - 138 = 74$ m (accept 70-80 m)
Question 12
A hiker walks $4$ km on a bearing of $060°$, then turns and walks $3$ km on a bearing of $150°$.
(a) Sketch the hiker's journey, showing both legs and the angle between them. (1 mark)
(b) Show that the angle between the two paths is $90°$. (1 mark)
(c) Calculate how far the hiker is from their starting point. (2 marks)
(d) On what bearing must the hiker walk to return directly to the start? (1 mark)
Marking Criteria
- 1 mark: Accurate sketch with North reference and bearings marked
- 1 mark: Angle between paths $= 150° - 60° = 90°$
- 1 mark: Apply Pythagoras: $d^2 = 4^2 + 3^2 = 25$
- 1 mark: Correct distance $d = 5$ km
- 1 mark: Correct back bearing calculation (accept $240°$ or reasoning about return direction)