AC9M10M03Year 10MeasurementPractised on site

AC9M10M03 — Year 10 Measurement

Official Curriculum Content Description:

solve practical problems applying Pythagoras’ theorem and trigonometry of right-angled triangles, including problems involving direction and angles of elevation and depression

Worked Examples (3)

Example 1Bearings & Navigation Geometry
Point B is located at an angle of 31∘31^\circ West of South from Point A. Calculate the three-figure bearing of B from A.
Common mistake: Measuring the acute angle directly instead of clockwise from North.
Correct Answer:211211
Worked Solution:
1. Bearings are measured clockwise from North as 3 digits: bearing = 211∘211^\circ.
Example 2Angles of Elevation & Depression
An observer stands 75 m75\text{ m} away from the base of a vertical tower. The angle of elevation to the top of the tower is 55∘55^\circ. Calculate the height of the tower hh. Give your answer to 1 decimal place.
TopObserverBase75 mh55°
Diagram not accurately drawn
Common mistake: Measuring the angle of elevation from the vertical tower line instead of the horizontal ground line.
Correct Answer:107.1 m107.1\text{ m}
Accepted: A unit is optional (m accepted).
Worked Solution:
1. The ground distance, tower height, and line of sight form a right-angled triangle. 2. tan⁡(55∘)=heightdistance=h75\tan(55^\circ) = \frac{\text{height}}{\text{distance}} = \frac{h}{75}. 3. h=75×tan⁡(55∘)≈107.1 mh = 75 \times \tan(55^\circ) \approx 107.1\text{ m}.
Example 3Trigonometric Ratios (Sine, Cosine, Tangent)
In a right-angled triangle, angle A=20∘A = 20^\circ and the hypotenuse is 14 cm14\text{ cm}. Find the length of the side adjacent to angle AA, xx. Give your answer to 1 decimal place.
ABC14 cmx20°
Diagram not accurately drawn
Common mistake: Fixing opposite/adjacent to the triangle orientation rather than angle AA, erroneously using sin⁡(20∘)\sin(20^\circ) instead of cos⁡(20∘)\cos(20^\circ).
Correct Answer:13.2 cm13.2\text{ cm}
Accepted: A unit is optional (cm accepted).
Worked Solution:
1. Identify the ratio: cos⁡A=adjacenthypotenuse\cos A = \frac{\text{adjacent}}{\text{hypotenuse}}. 2. Rearrange for the adjacent side: x=hypotenuse×cos⁡A=14×cos⁡(20∘)x = \text{hypotenuse} \times \cos A = 14 \times \cos(20^\circ). 3. x≈13.2 cmx \approx 13.2\text{ cm}.

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