SUBJECT: MATHEMATICS
CLASS: JSS 2
DATE:
TERM: 3rd TERM
REFERENCE
TOPIC: ANGLES OF ELEVATION AND DEPRESSION CONTENT: (i) Horizontal and vertical lines (ii) Angles of elevation (iii) Measuring angles of elevation and depression Horizontal and Vertical Lines Horizontal lines are lines that are parallel to the earth. For example, the surface of a liquid in a container, floor of a classroom, etc. Seethe 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. Evaluation: Reading Assignment NGM BK 2 Chapter 17, pg 173 Essential Mathematics for JSS BK 2, Chapter 17, pg 173 Angles of Elevation The angle of elevation of an object from a given point is the angle formed when an observer looks up to see an object his head. See the diagram below. angle of elevation Horizontal plane V = view point, T = top where the object is, F = foot of the vertical plane, e = angle of elevation Reading Assignment 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 the angle formed when an observer looks down to see an object below his head. Horizontal Thus the angle of elevation is equal in size to the angle of depression. (Alternate angles are equal i.e. d = e) NGM BK 2 Chapter 20, pgs 166 – 167 Measuring Angles of Elevation and Depression When constructing angles of elevation and depression, the use of scale drawing is necessary in order to have effective construction of angle. Consider the diagram below; find the height of the flagpole to the nearest metre using suitable scale. Solution By construction, choose a scale of 1cm represent 2m. The height of the flagpole PT = 3cm, converted to m, will give 2 x 3 = 6 Example 2: The angle of elevation of the top of a tower 42m away from a point on the level ground is 36o, find the height of the tower. Solution By construction, using a suitable scale of 1cm represented by 6cm, then PR = 426= 7cm The length TR = 5.0cm converting back to metre, we have; Length TR = 5 x 6 = 30m Example 3: From the top of a building 20m high, the angle of depression of a car is 45o, find the distance of the car from the foot of the building. Solution Rough sketch: 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 equal angle of elevation; By construction, using a suitable scale of 1cm represents 5m For 20m, we have 205 = 4cm Length CF 4cm By conversion, length CF = 4 x 5 = 20m EVALUATION GENERAL EVALUATION REVISION QUESTION READING ASSIGNMENT NGM BK 2 chapter 20, page 166 – 169 Essential mathematics for JSS BK 2, chapter 23, pg 295 – 297 Exercise 23.1 No 1, 2 & 3 page 296 WEEKEND ASSIGNMENT THEORY flagpole
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