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) Use the East and North components to find the bearing the hiker must walk to return directly to the start. Give the bearing to the nearest degree in three-digit notation. (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: Components give a net displacement of about 4.964 km East and 0.598 km South; the return vector is 4.964 km West and 0.598 km North. In the return right triangle, $\tan\theta = 4.964/0.598$, so $\theta \approx 83.1°$ west of North and the bearing is $360° - \theta \approx 276.9°$. Accept $277°$ to the nearest degree or a reasonable rounded equivalent.