Lesson Notes By Weeks and Term v4 - SHS 1
MEASUREMENT
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MEASUREMENT
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Subject: Mathematics
Class: SHS 1
Term: 2nd Term
Week: 4
Grade code: 1.3.2.LI.2
Strand code: 3
Sub-strand code: 2
Content standard code: 1.3.2.CS.21
Indicator code: 1.3.2.LI.2
Theme: GEOMETRY AROUND US
Subtheme: MEASUREMENT
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This lesson introduces students to the trigonometric ratios of special angles (30°, 45°, 60°) and the use of calculators for any angle. These concepts are not just for examinations; they are the foundation for many practical fields that shape our communities. For instance, when a carpenter builds a roof for a house in your community, they use angles to ensure the roof is steep enough for rainwater to run off. Surveyors use these same principles to map land for new roads or farms. By understanding these special angles, we gain a powerful tool for solving real-world problems involving triangles and measurements, without always needing a calculator.
A. Recap: The Basic Trigonometric Ratios (SOH CAH TOA)
Before we look at special angles, let's remember the basic definitions from JHS. For any right-angled triangle with an angle θ: Hypotenuse (H): The longest side, opposite the right angle. Opposite (O): The side directly opposite to the angle θ. Adjacent (A): The side next to the angle θ (that is not the hypotenuse).
The three main trigonometric ratios are: Sine: `sin(θ) = Opposite / Hypotenuse` (SOH) Cosine: `cos(θ) = Adjacent / Hypotenuse` (CAH) Tangent: `tan(θ) = Opposite / Adjacent` (TOA) B. Deriving Trig Ratios for the Special Angle 45°
We can find the exact values for sin(45°), cos(45°), and tan(45°) using a special triangle. Construct the Triangle: Start with a simple isosceles right-angled triangle. Since it's isosceles, the two shorter sides are equal. Let's make them 1 unit long. The angles in this triangle will be 90°, 45°, and 45°. Find the Hypotenuse: We use Pythagoras' theorem (a² + b² = c²). 1² + 1² = H² 1 + 1 = H² 2 = H² H = √2 Calculate the Ratios for 45°: Now we use SOH CAH TOA for one of the 45° angles. Opposite (O) = 1 Adjacent (A) = 1 Hypotenuse (H) = √2 sin(45°) = O/H = 1/√2. We rationalise the denominator by multiplying the top and bottom by √2: (1 * √2) / (√2 * √2) = √2 / 2 cos(45°) = A/H = 1/√2 = √2 / 2 tan(45°) = O/A = 1/1 = 1 C. Deriving Trig Ratios for the Special Angles 30° and 60°