Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Triagnoimetry ratios of special angles
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Triagnoimetry ratios of special angles
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Subject: Further Mathematics
Class: Senior Secondary 1
Term: 3rd Term
Week: 2
Theme: Pure Mathematics
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Derive trignometric ratios of special angles 30o,45o, 60o Apply the trignometric ratios of 30o, 45o, 60o to solve problems with out the use of tables
This section explains the methods for deriving the trigonometric ratios of 30°, 45°, and 60° and provides illustrative examples. The core idea is to use specific right-angled triangles where side lengths can be easily determined using basic geometry and Pythagoras' theorem. 2.
1. Basic Trigonometric Ratios (Recall) Before diving into special angles, a brief recall of SOH CAH TOA is essential for students: Sine (Sin) = Opposite / Hypotenuse Cosine (Cos) = Adjacent / Hypotenuse Tangent (Tan) = Opposite / Adjacent 2.
2. Derivation of Trigonometric Ratios for 45° To derive the ratios for 45°, consider an isosceles right-angled triangle. Let the two equal sides (opposite and adjacent to 45°) be of length 1 unit each. By Pythagoras' theorem, the hypotenuse (H) will be: H2 = 12 + 12 = 1 + 1 = 2 H = √2 units. Now, consider one of the 45° angles in this triangle: Opposite side = 1 Adjacent side = 1 Hypotenuse = √2 Therefore: sin 45° = Opposite / Hypotenuse = 1 / √2 = √2 / 2 (rationalizing the denominator) cos 45° = Adjacent / Hypotenuse = 1 / √2 = √2 / 2 tan 45° = Opposite / Adjacent = 1 / 1 = 1 Diagram for 45°: ``` A |\ | \ √2 1 | \ | \ |____\ B 1 C (Angle B = 90°, Angle A = 45°, Angle C = 45°) ``` 2.
3. Derivation of Trigonometric Ratios for 30° and 60° To derive the ratios for 30° and 60°, consider an equilateral triangle. Let each side of the equilateral triangle be 2 units long. All angles are 60°. Draw an altitude (height) from one vertex to the midpoint of the opposite side. This altitude bisects the vertex angle and the base. This creates two congruent 30°-60°-90° right-angled triangles. Consider one of these right-angled triangles: Hypotenuse = 2 units (original side of equilateral triangle) Adjacent side to 60° (or opposite side to 30°) = 1 unit (half of the base) Using Pythagoras' theorem to find the third side (altitude, opposite to 60° and adjacent to 30°): (Altitude)2 + 12 = 22 (Altitude)2 + 1 = 4 (Altitude)2 = 3 Altitude = √3 units. Diagram for 30° and 60°: ``` A /|\ / | \ / | \ 2 / |√3 \ 2 / | \ /_____|_____\ C' 1 D 1 B' (Equilateral Triangle ABC, D is midpoint of CB. Triangle ADC is 30-60-90) (Angle C = 60°, Angle ADC = 90°, Angle CAD = 30°) ``` Now, calculate the ratios for 30°: Opposite side = 1 Adjacent side = √3 Hypotenuse = 2 Therefore: sin 30° = Opposite / Hypotenuse = 1 / 2 cos 30° = Adjacent / Hypotenuse = √3 / 2 tan 30° = Opposite / Adjacent = 1 / √3 = √3 / 3 Now, calculate the ratios for 60°: Opposite side = √3 Adjacent side = 1 Hypotenuse = 2 Therefore: sin 60° = Opposite / Hypotenuse = √3 / 2 cos 60° = Adjacent / Hypotenuse = 1 / 2 tan 60° = Opposite / Adjacent = √3 / 1 = √3 2.
4. Summary Table of Special Angle Ratios Teachers should guide students to construct and memorise this table. | Angle (θ) | sin θ | cos θ | tan θ | | :-------- | :--------- | :--------- | :--------- | | 30° | 1/2 | √3 / 2 | √3 / 3 | | 45° | √2 / 2 | √2 / 2 | 1 | | 60° | √3 / 2 | 1/2 | √3 | 2.
5. Worked
Examples: Example 1: Evaluating Expressions Evaluate without using tables or a calculator: sin 60° cos 30° + tan 45°.
Step 1: Identify the values from the summary table. sin 60° = √3 / 2 cos 30° = √3 / 2 tan 45° = 1 Step 2: Substitute the values into the expression. (√3 / 2) (√3 / 2) + 1 * *Step 3: Perform 45° | √2 / 2 | √2 / 2 | 1 | | 60° | √3 / 2 | 1/2 | √3 | 2.
5. Worked
Examples: Example 1: Evaluating Expressions Evaluate without using tables or a calculator: sin 60° cos 30° + tan 45°.
Step 1: Identify the values from the summary table. sin 60° = √3 / 2 cos 30° = √3 / 2 tan 45° = 1 Step 2: Substitute the values into the expression. (√3 / 2) (√3 / 2) + 1 Step 3: Perform the multiplication. (√3 √3) / (2 2) + 1 = 3 / 4 + 1 Step 4: Perform the addition. 3 / 4 + 4 / 4 = 7 / 4 or 1 3⁄4 Example 2: Solving for an Unknown in a Right-Angled Triangle A surveyor in Enugu sights the top of a newly installed telecom mast at an angle of elevation of 60°. If the surveyor is standing 50 metres from the base of the mast, calculate the height of the mast (h) to the nearest metre.
Step 1: Draw a diagram representing the situation. (A right-angled triangle with angle 60°, adjacent side 50m, opposite side h)
Step 2: Identify the trigonometric ratio that relates the given angle, the known side (adjacent), and the unknown side (opposite).
This is the tangent ratio: tan θ = Opposite / Adjacent.
Step 3: Substitute the known values and the special angle ratio. tan 60° = h / 50 We know tan 60° = √
3. So, √3 = h / 50 Step 4: Solve for h. h = 50 √3 Using √3 ≈ 1.732 (students can be provided with this value or asked to use it if calculator is allowed for final computation, otherwise leave in surd form) h = 50 1.732 = 86.6 metres.
Step 5: Round to the nearest metre as required. h ≈ 87 metres.
Example 3: More Complex Evaluation Evaluate: (sin 30° + cos 60°) / tan 45° Step 1: Identify the values. sin 30° = 1/2 cos 60° = 1/2 tan 45° = 1 Step 2: Substitute into the expression. (1/2 + 1/2) / 1 * Step 3: Perform the operations. (1) / 1 = 1 This section outlines activities for both the teacher and the students to ensure an interactive and effective learning experience. 3.
1. Introduction (10-15 minutes)
Teacher Activity: Initiate a quick recap of basic trigonometric ratios (SOH CAH TOA) and their application in right-angled triangles using a simple problem (e.g., finding a side given an angle and another side, using a calculator for general angles). Introduce the concept of "special angles" (30°, 45°, 60°) and explain why they are significant (exact values, no calculator needed).
Pose a challenge: "How can we find the exact values of sin 30° or tan 60° without a calculator?" Student Activity: Actively participate in the recap, answering questions on SOH CAH TOA. Engage in the discussion about special angles. Consider the challenge posed by the teacher. 3.
2. Derivation of 45° Ratios (15-20 minutes)
Teacher Activity: Guide students to draw an isosceles right-angled triangle on their exercise books or on a board. Suggest using sides of 1 unit or any convenient equal length. Prompt students to apply Pythagoras' theorem to find the hypotenuse. Facilitate the calculation of sin 45°, cos 45°, and tan 45° based on the derived side lengths. Emphasize rationalising denominators where necessary.
Student Activity: Draw the isosceles right-angled triangle. Calculate the hypotenuse. In pairs or small groups, derive the sin, cos, and tan ratios for 45°. Share their derivations with the class. 3.
3. Derivation of 30° and 60° Ratios (20-25 minutes)
Teacher Activity: Instruct students to draw an equilateral triangle (e.g., with side lengths of 2 units). Guide them to draw an altitude from one vertex to the midpoint of the opposite side, explaining that this creates a 30°-60°-90° triangle. Prompt students to apply Pythagoras' theorem to find the length of the altitude. Facilitate the calculation of sin 30°, cos 30°, tan 30°, and then sin 60°, cos 60°, tan 60° from this single right-angled triangle. Encourage students to articulate the relationships between the sides and angles.
Student Activity: Draw an equilateral triangle and its altitude. Calculate the length of the altitude. In pairs or small groups, derive all six trigonometric ratios for 30° and 60°. Compare and discuss their results. 3.
4. Consolidation and Summary (5-10 minutes)
Teacher Activity: Lead a class discussion to compile all derived ratios into a summary table on the board. Highlight any patterns or relationships (e.g., sin 30° = cos 60°). Advise students on memorization techniques (e.g., using the "finger trick" or just understanding the derivation).
Student Activity: Copy the summary table into their notebooks. Ask clarifying questions. Engage in discussions about patterns observed. 3.
5. Problem Solving and Application (20-25 minutes)
Teacher Activity: Present the worked examples (from Key Concepts and Explanations section) on the board, explaining each step clearly. Emphasize the importance of substituting exact values (fractions and surds) and simplifying. Provide Guided Practice questions for students to work on immediately. Circulate, providing support and checking understanding.
Student Activity: Follow along with the teacher's examples, taking notes. Attempt Guided Practice questions individually or in pairs. Share their solutions and reasoning with the class. The teacher should provide these questions for students to solve in class, offering support and guidance as needed.
Question 1: Evaluate without using tables or a calculator: 2 sin 30° + cos 60° - tan 45°.
Solution 1: Step 1: Recall the values of the ratios. sin 30° = 1/2 cos 60° = 1/2 tan 45° = 1 Step 2: Substitute the values into the expression. 2(1/2) + (1/2) - 1 Step 3: Perform the operations. 1 + 1/2 - 1 = 1/2
Commentary: This question tests basic substitution and arithmetic with fractions. It reinforces the direct recall of the derived ratios.
Question 2: Find the value of `x` in the right-angled triangle below. (The teacher should draw a right-angled triangle with angles 90°, 30°, and 60°. Label the side opposite 60° as `x` and the side adjacent to 60° as 8 cm).
Solution 2: Step 1: Identify the knowns and unknowns in relation to the given angle (60°). Angle = 60° Opposite side = `x` Adjacent side = 8 cm Step 2: Choose the appropriate trigonometric ratio. The ratio relating Opposite and Adjacent is Tangent. tan θ = Opposite / Adjacent Step 3: Substitute the values. tan 60° = x / 8 Step 4: Substitute the special angle value for tan 60°. √3 = x / 8 Step 5: Solve for `x`. x = 8√3 cm
Commentary: This problem applies the special angle ratios in a geometric context, requiring students to identify sides relative to a given angle and select the correct ratio. Keeping the answer in surd form is crucial when not using a calculator.
Question 3: Simplify: (tan2 60° - sin2 45°) / (cos 30° sin 60°)
Solution 3: Step 1: Recall the values of the ratios. tan 60° = √3 sin 45° = √2 / 2 cos 30° = √3 / 2 sin 60° = √3 / 2 Step 2: Substitute the values into the expression. [ (√3)2 - (√2 / 2)2 ] / [ (√3 / 2) * (√3 / 2) ] Step 3: Evaluate the squares and the product. ( 3 - 2/4 ) / ( 3/4 ) ( 3 - 1/2 ) / ( 3/4 )
Step 4: Simplify the numerator. ( 6/2 - 1/2 ) / ( 3/4 ) = ( 5/2 ) / ( 3/4 )
Step 5: Perform the division by multiplying by the reciprocal. (5/2) (4/3) = (5 4) / (2 * 3) = 20 / 6 = 10 / 3
Commentary: This question involves multiple operations and powers, testing students' proficiency in both ratio substitution and algebraic manipulation of fractions and surds.
Construction of Ramps and Staircases: In Nigeria, ramps for wheelchair access or staircases in multi-story buildings often need specific angles for safety and usability. Special angles like 30° or 45° are frequently used as design benchmarks. Students can calculate the exact length of the ramp surface or the vertical rise for a given horizontal distance without a calculator if these angles are applied. For example, ensuring a ramp for a hospital in Kano meets specific accessibility codes requiring a 30° incline.
Land Surveying and Mapping: Surveyors in rural and urban areas of Nigeria use trigonometry to determine property boundaries, measure land area, and create topographical maps. When mapping a new road in a hilly area like Jos, they might use the angle of elevation (e.g., 60°) to calculate the height of a hill or a point of interest from a known horizontal distance, often relying on special angle ratios for quick, on-the-spot estimations.
Engineering Design of Structures: Civil and mechanical engineers in Nigeria frequently use trigonometry to design various structures like communication towers, bridges, and electrical transmission poles. For instance, determining the optimal length and angle of guy wires to support a tall mast, or calculating forces on structural components, often involves resolving vectors using these special trigonometric ratios. This ensures the stability and safety of infrastructure projects across the country.