Lesson Notes By Weeks and Term v3 - Junior Secondary 2
Angles
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Angles
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Subject: General Mathematics
Class: Junior Secondary 2
Term: 3rd Term
Week: 4
Theme: Mensuration And Geometry
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Find the sum of angles of triangles Find sum of angles in a quadrilateral Find the sum of the in terior angles of a polygon Distinguish between angles of elevation and angles of depression; Use angles of elevation and depression in calculating distances and heights using scale drawing Identify the relationship between angles of elevation and depression Solve problems on quantitative aptitude related to angles.
Mensuration And Geometry Angles Term: 3rd Term Week: 11 ---
1. Overview and Learning Objectives This lesson introduces students to advanced concepts of angles, building upon their prior knowledge of basic angle types and relationships. The study of angles is fundamental in various fields, including engineering, architecture, surveying, navigation, and even sports. Understanding how angles behave within geometric shapes and in relation to an observer's line of sight provides a crucial foundation for problem-solving in real-world scenarios in Nigeria, such as constructing houses, designing bridges, surveying land for agriculture, or even calculating distances when observing objects like a telecommunication mast or a distant building. Upon completion of this lesson, students will be able to: Determine the sum of interior angles in any triangle. Determine the sum of interior angles in any quadrilateral. Calculate the sum of the interior angles of any convex polygon. Differentiate between angles formed when looking upwards (angle of elevation) and downwards (angle of depression). Apply the concepts of angles of elevation and depression to calculate heights and distances using scale drawings. Recognise and explain the relationship between an angle of elevation and a corresponding angle of depression. * Solve problems requiring quick application of angle properties, often encountered in quantitative aptitude tests.
2. Key Concepts and Explanations 2.
1. Sum of Angles in a Triangle A triangle is a polygon with three sides and three interior angles.
Concept: The sum of the interior angles of any triangle is always 180 degrees.
Explanation/Proof:
1. Consider a triangle ABC.
2. Draw a line XY parallel to the base BC passing through the vertex A.
3. Angle XAB and Angle ABC are alternate interior angles (because XY || BC and AB is a transversal), so $\angle XAB = \angle ABC$.
4. Angle YAC and Angle ACB are alternate interior angles (because XY || BC and AC is a transversal), so $\angle YAC = \angle ACB$.
5. Angles XAB, BAC, and YAC form a straight line angle at vertex A. The sum of angles on a straight line is 180°.
6. Therefore, $\angle XAB + \angle BAC + \angle YAC = 180°$.
7. Substituting the alternate angles: $\angle ABC + \angle BAC + \angle ACB = 180°$.
Conclusion: Sum of angles of a triangle = 180°. Worked Example 2.1.1: In a triangle PQR, $\angle P = 65°$ and $\angle Q = 50°$. Find $\angle R$.
Solution: Let $\angle R = x$. Sum of angles in a triangle = 180° $\angle P + \angle Q + \angle R = 180°$ $65° + 50° + x = 180°$ $115° + x = 180°$ $x = 180° - 115°$ $x = 65°$ Therefore, $\angle R = 65°$. 2.
2. Sum of Angles in a Quadrilateral A quadrilateral is a polygon with four sides and four interior angles.
Concept: The sum of the interior angles of any quadrilateral is always 360 degrees.
Explanation/Proof:
1. Consider a quadrilateral ABCD.
2. Draw a diagonal from one vertex to an opposite vertex (e.g., AC).
3. This diagonal divides the quadrilateral into two triangles: $\triangle ABC$ and $\triangle ADC$.
4. The sum of angles in $\triangle ABC = 180°$.
5. The sum of angles in $\triangle ADC = 180°$.
6. The sum of the angles of the quadrilateral is the sum of the angles of these two triangles.
7. Therefore, Sum of angles in a quadrilateral = $180° + 180° = 360°$. Worked Example 2.2.1: A plot of land in Abuja has a quadrilateral shape with angles $90°$, $110°$, $75°$, and $x$. Find the value of $x$.
Solution: Sum of angles in a quadrilateral = 360° $90° + 110° + 75° + x = 360°$ $275° + x = 360°$ $x = 360° - 275°$ $x = 85°$ The fourth angle is $85°$. 2.
3. Sum of Interior Angles of a Polygon A polygon is a closed plane figure bounded by straight line segments. A convex polygon has all its interior angles less than 180°.
Concept: The sum of the interior angles of a convex polygon with 'n' sides is given by the formula: $(n - 2) \times 180°$.
Explanation/Derivation:
1. Any convex polygon with 'n' sides can be divided + 110° + 75° + x = 360°$ $275° + x = 360°$ $x = 360° - 275°$ $x = 85°$ The fourth angle is $85°$. 2.
3. Sum of Interior Angles of a Polygon A polygon is a closed plane figure bounded by straight line segments. A convex polygon has all its interior angles less than 180°.
Concept: The sum of the interior angles of a convex polygon with 'n' sides is given by the formula: $(n - 2) \times 180°$.
Explanation/Derivation:
1. Any convex polygon with 'n' sides can be divided into $(n - 2)$ non-overlapping triangles by drawing diagonals from a single vertex.
2. For example: A triangle (n=3) can be divided into (3-2) = 1 triangle. Sum = $1 \times 180° = 180°$. A quadrilateral (n=4) can be divided into (4-2) = 2 triangles. Sum = $2 \times 180° = 360°$. A pentagon (n=5) can be divided into (5-2) = 3 triangles. Sum = $3 \times 180° = 540°$.
3. Since the sum of angles in each triangle is 180°, the total sum of interior angles of the polygon is $(n - 2)$ times 180°. Worked Example 2.3.1: Calculate the sum of the interior angles of a hexagon.
Solution: A hexagon has $n = 6$ sides. Sum of interior angles = $(n - 2) \times 180°$ $= (6 - 2) \times 180°$ $= 4 \times 180°$ $= 720°$ The sum of the interior angles of a hexagon is $720°$. Worked Example 2.3.2: Each interior angle of a regular polygon is $144°$. Find the number of sides of the polygon.
Solution: Let the number of sides be $n$. Sum of interior angles = $(n - 2) \times 180°$. For a regular polygon, all interior angles are equal. So, each interior angle = $\frac{(n - 2) \times 180°}{n}$. Given that each interior angle is $144°$: $144 = \frac{(n - 2) \times 180}{n}$ $144n = 180n - 360$ $360 = 180n - 144n$ $360 = 36n$ $n = \frac{360}{36}$ $n = 10$ The polygon has 10 sides (a decagon). Alternative method (using exterior angles): Each exterior angle = $180° - 144° = 36°$. Sum of exterior angles of any convex polygon is $360°$. Number of sides = $\frac{360°}{\text{each exterior angle}} = \frac{360°}{36°} = 10$. 2.
4. Angles of Elevation and Depression These angles are used in trigonometry to describe the relationship between an observer and an object, relative to a horizontal line. 2.4.
1. Horizontal Line: This is an imaginary line that extends straight out from the observer's eye, parallel to the ground. 2.4.
2. Line of Sight: This is the imaginary line connecting the observer's eye to the object being viewed.
Angle of Elevation: When an observer looks up at an object, the angle formed between the horizontal line and the line of sight is called the angle of elevation. Imagine looking up at a kite in the sky or the top of a tall building in Lagos.
Angle of Depression: When an observer looks down at an object, the angle formed between the horizontal line (from the observer) and the line of sight is called the angle of depression. Imagine looking down from a high-rise building at a car on the street below or from an aircraft at a landmark. Relationship between Angle of Elevation and Depression: If an observer at point A looks up at an object at point B, the angle of elevation from A to B is formed. If an observer at point B (where the object is) looks down at the original observer at point A, the angle of depression from B to A is formed. These two angles (angle of elevation from A to B and angle of depression from B to A) are equal. This is because the two horizontal lines (one from A and one from B) are parallel, and the line of sight AB acts as a transversal. Thus, they are alternate interior angles.
Diagram Illustration: ``` Observer's Horizontal Line (B) B----------------------------------- | \ Angle of Depression (from B to A) | \ | \ Line of Sight | \ | \ | \ | of depression from B to A is formed. These two angles (angle of elevation from A to B and angle of depression from B to A) are equal. This is because the two horizontal lines (one from A and one from B) are parallel, and the line of sight AB acts as a transversal. Thus, they are alternate interior angles.
Diagram Illustration: ``` Observer's Horizontal Line (B) B----------------------------------- | \ Angle of Depression (from B to A) | \ | \ Line of Sight | \ | \ | \ | \ | \ Ground Level ------- A ---------------- Object's Horizontal Line (A) Angle of Elevation (from A to B) ``` If a person at A looks at B, the angle of elevation is formed at A. If a person at B looks at A, the angle of depression is formed at B. These two angles are equal (alternate angles). 2.
5. Using Angles of Elevation and Depression with Scale Drawing Scale drawing is a method of representing actual objects or areas using a specific ratio (scale). It allows us to calculate actual distances and heights by measuring on the drawing. Steps for solving problems using scale drawing:
1. Read the problem carefully: Identify known distances, heights, and angles.
2. Choose an appropriate scale: A scale like 1 cm to 5 m or 1:500 means 1 cm on the drawing represents 5 meters or 500 cm in reality. The scale should allow the drawing to fit on the page while being large enough for accurate measurements.
3. Draw the horizontal line: This represents the ground level or the observer's horizontal line of sight.
4. Mark the observer's position: Place a point on the horizontal line for the observer.
5. Draw the angle of elevation/depression: Using a protractor, draw the given angle from the observer's position relative to the horizontal line.
6. Draw the line of sight: Extend the line from the angle.
7. Draw the object: If the height is known, mark it along a vertical line from the ground. If the distance is known, mark it along the horizontal line. Complete the right-angled triangle.
8. Measure the unknown: Use a ruler to measure the required distance or height on your drawing.
9. Convert to actual dimensions: Multiply the measured length on the drawing by your chosen scale factor to get the real-life measurement. Worked Example 2.5.1: A surveyor stands 20 meters from the base of a telecommunication mast in Enugu. He measures the angle of elevation to the top of the mast as $40°$. Using a scale drawing, estimate the height of the mast.
Solution:
1. Choose a scale: Let's use 1 cm represents 5 meters (Scale 1:500). Distance from mast = 20 m = $\frac{20}{5}$ cm = 4 cm on drawing.
2. Draw a horizontal line: Represents the ground. Mark point A for the surveyor.
3. Measure distance: From A, measure 4 cm horizontally to point B (base of mast).
4. Draw vertical line: From B, draw a vertical line upwards (representing the mast).
5. Draw angle of elevation: At point A, using a protractor, draw an angle of $40°$ from the horizontal line (AB) upwards.
6. Draw line of sight: Extend this line until it intersects the vertical line from B at point C (top of mast).
7. Measure height: Measure the length of the line segment BC on your drawing. (Expected measurement if drawn accurately): BC should be approximately 3.36 cm.
8. Convert to actual height: Actual height = Measured length on drawing $\times$ Scale factor Actual height $\approx 3.36 \text{ cm} \times 5 \text{ m/cm} = 16.8 \text{ m}$.
Therefore, the estimated height of the mast is approximately 16.8 meters. (Note for teacher: Encourage students to use accurate drawing instruments. Small variations are expected due to manual measurement.) 2.
6. Quantitative Aptitude Related to Angles Quantitative aptitude problems on angles usually involve applying the properties of angles (sum of angles in a triangle/quadrilateral/polygon, angles on a straight line, vertically opposite angles, parallel lines properties) to quickly find missing angles or relate angles in complex diagrams. These problems test understanding and speed