SUBJECT: MATHEMATICS
CLASS: JSS 3
DATE:
TERM: 2nd TERM
TOPIC: ANGLES OF ELEVATION AND DEPRESSION CONTENT: i) Horizontal and vertical lines iii) Measuring angles of elevation and depression. Horizontal and Vertical Lines Horizontal lines are lines that are parallel to the surface of the earth. For example, the surface of a liquid in a container, floor of a classroom, etc. See the diagram below: Horizontal line Vertical lines are lines that are perpendicular to the horizontal surface, e.g. wall of a classroom, a swing pendulum, etc. Vertical line Evaluation: Say whether the following are horizontal or vertical or neither. READING ASSIGNMENT NGM Bk 2 Chapter 20, Page 165 Essential Mathematics for JSS Bk 2, Chapter 17, Pg 173 Angles of Elevation The angle of elevation of an object for a given point above the surface of the earth is the angle formed between the horizontal plane and the view point of the object. See the diagram below. T T V F VF Horizontal plane V = View point, T = Top where the object is, F = Foot of the vertical plane e = angle of elevation Reference: NGM Bk 2 Chapter 20, Page 165 Essential Mathematics for JSS Bk 2, Chapter 17, Pg 173 Angle of Depression The angle of depression of an object from a given point T is angle from the horizontal line above the earth’s surface and the vertical surface. Horizontal T h Thus, the angle of elevation is equal in size to the angle of depression. (alternate angles are equal i.e. d = e) Measuring Angles of Elevation and Depression When constructing angles of elevation and depression, the use of scale drawings is necessary in order to have effective construction of angles Worked Examples scale. F Q 10m T Solution By construction, choose a scale of 1cm represent 2m. The height of the flagpole PT = 3cm, converted to m, will give 2 × 3 = 6. Example 2: The angle of elevation of the top of a tower to a point 42m away from its base on level ground is 360, find the height of the tower. Solution T 3 R 42m By construction, using a suitable scale of 1cm represented by 6m, then PR = 42/6 = 7cm. The length TR = 5.0cm. Converting back to metre, we have; Length TR = 5 × 6 = 30m Example 3: From the top of a bulding 20m high, the angle of depression of a car is 450, find the distance of the car from the foot of the building. Solution Rough sketch: T C 20m F T = Top of the building, C = Car, F = Foot of the building CF is the distance of the car from the foot of the building. Since angle of depression equals angle of elevation; By construction, using a suitable scale of 1cm represents 5m For 20m, we have 20/5 = 4cm Length CF = 4cm By conversion, length CF = 4 × 5 = 20m. Evaluation: angle of elevation of Q from R. Q 10m P 50m R sea is 350, find the distance between the boat and the foot of the cliff. READING ASSIGNMENT NGM Bk 2 Chapter 20, Page 166 - 169 Essential Mathematics for JSS Bk 2, Chapter 17, Pg 176 – 177. WEEKEND ASSIGNMENT Objective THEORY F 20cm B C 8cm
WEEK EIGHT Date:…………………….
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