Lesson Notes By Weeks and Term v3 - Junior Secondary 3
Trigonometry
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Trigonometry
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Subject: General Mathematics
Class: Junior Secondary 3
Term: 2nd Term
Week: 4
Theme: Mensuration And Geometry
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Identify sine, cosine and tangent of an acute angle Solve problems on applications of trigonometric ratios to finding distances and lengths Apply trigonometric ratios in solving world problems.
at this level for trigonometric values.
Worked Example 2: Finding an Unknown Side A ladder leaning against a wall makes an angle of 60° with the ground. If the base of the ladder is 2.5 meters away from the wall, how high up the wall does the ladder reach? (Assume the wall is perpendicular to the ground).
Solution 2:
1. Sketch the diagram: Wall | \ | \ h | \ L (ladder) | \ |_____\ Ground 60°
2. Identify knowns and unknowns relative to the given angle (60°): Angle $\theta = 60^\circ$ Distance from wall (Adjacent) = 2.5 m (Known) Height up the wall (Opposite) = h (Unknown) Hypotenuse (L, length of ladder) is not given and not directly asked for.
3. Select the appropriate ratio: We have the Opposite (h) and the Adjacent (2.5 m). The ratio that relates Opposite and Adjacent is Tangent (TOA). $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$
4. Formulate and solve the equation: $\tan(60^\circ) = \frac{h}{2.5}$ From a calculator, $\tan(60^\circ) \approx 1.732$ $1.732 = \frac{h}{2.5}$ $h = 1.732 \times 2.5$ $h = 4.33$ meters (to 2 decimal places) Therefore, the ladder reaches approximately 4.33 meters up the wall.
Worked Example 3: Finding another Unknown Side A kite string is 50 meters long and makes an angle of 35° with the ground. Assuming the string is taut, what is the height of the kite above the ground?
Solution 3:
1. Sketch the diagram: Kite |\ | \ 50m (Hypotenuse) h | \ | \ |____\ Ground 35°
2. Identify knowns and unknowns relative to the given angle (35°): Angle $\theta = 35^\circ$ Kite string length (Hypotenuse) = 50 m (Known) Height of the kite (Opposite) = h (Unknown) Adjacent side (distance on ground) is not given and not asked for.
3. Select the appropriate ratio: We have the Opposite (h) and the Hypotenuse (50 m). The ratio that relates Opposite and Hypotenuse is Sine (SOH). $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$
4. Formulate and solve the equation: $\sin(35^\circ) = \frac{h}{50}$ From a calculator, $\sin(35^\circ) \approx 0.5736$ $0.5736 = \frac{h}{50}$ $h = 0.5736 \times 50$ * $h = 28.68$ meters (to 2 decimal places) Therefore, the kite is approximately 28.68 meters above the ground.
A. Right-Angled Triangles: Revisiting Fundamentals A right-angled triangle is a triangle that has one angle exactly equal to 90 degrees. The side opposite the 90-degree angle is the longest side and is called the hypotenuse. The other two angles in a right-angled triangle are always acute angles (less than 90 degrees) and sum up to 90 degrees. For any given acute angle in a right-angled triangle, the other two sides are named relative to that angle: Hypotenuse (H): The side opposite the right angle. It is always the longest side.
Opposite (O): The side directly opposite the acute angle under consideration.
Adjacent (A): The side next to the acute angle under consideration, but not the hypotenuse.
Illustration: Consider a right-angled triangle ABC, with the right angle at
B. C |\ | \ | \ Hypotenuse | \ | \ A-----B (If considering angle C): Side AB is Opposite to angle
C. Side BC is Adjacent to angle
C. Side AC is the Hypotenuse. (If considering angle A): Side BC is Opposite to angle A. Side AB is Adjacent to angle A. Side AC is the Hypotenuse. It is crucial for students to correctly identify these sides relative to the specific acute angle being used in a problem. B. Trigonometric Ratios (SOH CAH TOA) Trigonometric ratios are constant ratios of the sides of a right-angled triangle for a given acute angle.
There are three primary ratios: Sine (sin), Cosine (cos), and Tangent (tan). These can be remembered using the mnemonic SOH CAH TOA: SOH: Sine = Opposite / Hypotenuse $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$ CAH: Cosine = Adjacent / Hypotenuse $\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$ TOA: Tangent = Opposite / Adjacent $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$ Where $\theta$ (theta) represents the acute angle.
Worked Example 1: Identifying Sides and Stating Ratios Consider a right-angled triangle PQR, right-angled at Q. Let PQ = 3 cm, QR = 4 cm, and PR = 5 cm. P |\ | \ 5 cm | \ 3 cm| \ | \ Q-----R 4 cm a) Identify the opposite, adjacent, and hypotenuse relative to angle P. b) Identify the opposite, adjacent, and hypotenuse relative to angle R. c) State the values of $\sin P$, $\cos P$, $\tan P$. d) State the values of $\sin R$, $\cos R$, $\tan R$.
Solution 1: a)
Relative to angle P: Opposite (O) = QR = 4 cm Adjacent (A) = PQ = 3 cm Hypotenuse (H) = PR = 5 cm b)
Relative to angle R: Opposite (O) = PQ = 3 cm Adjacent (A) = QR = 4 cm Hypotenuse (H) = PR = 5 cm c)
Values for angle P: $\sin P = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{4}{5}$ $\cos P = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{3}{5}$ $\tan P = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QR}{PQ} = \frac{4}{3}$ d)
Values for angle R: $\sin R = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{3}{5}$ $\cos R = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{4}{5}$ $\tan R = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{PQ}{QR} = \frac{3}{4}$ C. Using Trigonometric Ratios to Find Unknown Sides To find an unknown side of a right-angled triangle, students need to:
1. Identify the known angle.
2. Label the known side and the unknown side relative to the known angle (Opposite, Adjacent, or Hypotenuse).
3. Select the appropriate trigonometric ratio (SOH CAH TOA) that involves the known side, the unknown side, and the known angle.
4. Formulate the equation and solve for the unknown side.
Note: Students will require a scientific calculator for this part, as NERDC curriculum often expects calculator use at this level for trigonometric values.
Worked Example 2: Finding an Unknown Side A ladder leaning against a wall makes an angle of 60° with the ground. If the base of the ladder is 2.5 meters away from the wall, how high up the wall does the ladder reach? (Assume the wall is perpendicular to the ground).
Solution 2:
1. Sketch the diagram: Wall | \ | \ h | \ L (ladder) | \ |_____\ Ground 60°
2. Identify knowns and unknowns relative to the given angle (60°): Angle $\theta
A. Introduction (5-10 minutes)
Review Prior Knowledge: The teacher initiates a brief Q&A session on types of triangles, focusing on right-angled triangles and the Pythagoras theorem. Ask students to identify the hypotenuse in various right-angled triangles.
Introduce Trigonometry: The teacher explains that while Pythagoras helps with sides, trigonometry helps relate sides and angles.
Pose a scenario: "How can we find the height of the school flagpole without climbing it or measuring its shadow?" This sets the context for trigonometry.
B. Development of Key Concepts (25-30 minutes)
Identifying Sides: The teacher draws various right-angled triangles on the board, marking one acute angle in each. The teacher models how to correctly identify the Hypotenuse, Opposite, and Adjacent sides relative to the marked angle.
Student Activity: Students draw their own right-angled triangles, mark an angle, and label the sides. The teacher circulates to check understanding. Introducing Trigonometric Ratios (SOH CAH TOA): The teacher formally introduces Sine, Cosine, and Tangent as ratios of sides. The teacher writes SOH CAH TOA on the board and explains its meaning, emphasizing memorization. Using the triangles from the previous activity, the teacher demonstrates how to write the ratios (e.g., $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$) using the labelled sides.
Student Activity: Students practice writing the three ratios for different marked angles in triangles provided by the teacher, given numerical side lengths (e.g., from Worked Example 1). Using Calculators for Trigonometric Values: The teacher briefly demonstrates how to use a scientific calculator to find the sine, cosine, or tangent of a given angle (e.g., $\sin 30^\circ$, $\cos 45^\circ$, $\tan 60^\circ$). Emphasize ensuring the calculator is in "DEG" (degrees) mode.
Student Activity: Students practice finding values for a few angles using their calculators.
C. Application to Problem Solving (30-35 minutes)
Finding Unknown Sides - Teacher Led: The teacher presents a problem (like Worked Example 2 or 3) on the board.
Step 1: Guide students to sketch the problem diagram.
Step 2: Guide students to label the known angle, the known side, and the unknown side (O, A, or H).
Step 3: Guide students to choose the correct trigonometric ratio (SOH CAH TOA) that links the known angle, the known side, and the unknown side.
Step 4: Set up the equation and solve for the unknown, demonstrating calculator use.
Guided Practice: The teacher provides 2-3 similar problems for students to attempt in pairs or small groups. The teacher circulates, offering assistance and correcting misconceptions. Selected students present their solutions on the board, explaining their steps.
D. Conclusion (5 minutes)
Recap: The teacher quickly reviews the main points: identifying sides, the SOH CAH TOA mnemonic, and how to use ratios to find unknown lengths.
Homework Assignment: Assign independent practice questions.
Question 1: In the right-angled triangle XYZ, right-angled at Y, XY = 8 cm, YZ = 6 cm, and XZ = 10 cm. a) For angle X, identify the Opposite, Adjacent, and Hypotenuse sides. b) Calculate $\sin X$, $\cos X$, and $\tan X$.
Solution 1: a)
Diagram: X |\ | \ 10 cm 8 | \ cm| \ | \ Y-----Z 6 cm Relative to angle X: Opposite (O) = YZ = 6 cm Adjacent (A) = XY = 8 cm Hypotenuse (H) = XZ = 10 cm b)
Calculations: $\sin X = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{YZ}{XZ} = \frac{6}{10} = \frac{3}{5}$ $\cos X = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{XY}{XZ} = \frac{8}{10} = \frac{4}{5}$ $\tan X = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{YZ}{XY} = \frac{6}{8} = \frac{3}{4}$ Question 2: A ramp is built for a wheelchair user. The ramp makes an angle of 15° with the ground. If the horizontal distance covered by the ramp is 4 meters, what is the length of the ramp (hypotenuse)?
Solution 2: Diagram: Ramp (L) |\ | \ h | \ | \ |____\ Ground 15° Identify knowns and unknowns relative to 15°: Angle $\theta = 15^\circ$ Horizontal distance (Adjacent) = 4 m (Known) Length of ramp (Hypotenuse) = L (Unknown)
Select ratio: We have Adjacent and want Hypotenuse. This is Cosine (CAH). $\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$ Formulate and solve: $\cos(15^\circ) = \frac{4}{L}$ From calculator, $\cos(15^\circ) \approx 0.9659$ $0.9659 = \frac{4}{L}$ $L = \frac{4}{0.9659}$ $L \approx 4.14$ meters (to 2 decimal places)
Commentary: The length of the ramp is approximately 4.14 meters. This calculation is useful for engineers and builders to ensure accessibility standards are met.
Question 3: A student standing 20 meters away from the base of a cocoa tree observes the top of the tree at an angle of 40° to the horizontal. Calculate the height of the cocoa tree.
Solution 3: Diagram: Tree Top |\ | \ Height| \ (h) | \ |____\ Student 40° Identify knowns and unknowns relative to 40°: Angle $\theta = 40^\circ$ Distance from tree (Adjacent) = 20 m (Known) Height of tree (Opposite) = h (Unknown)
Select ratio: We have Opposite and Adjacent. This is Tangent (TOA). $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$ Formulate and solve: $\tan(40^\circ) = \frac{h}{20}$ From calculator, $\tan(40^\circ) \approx 0.8391$ $0.8391 = \frac{h}{20}$ $h = 0.8391 \times 20$ $h = 16.782$ meters $h \approx 16.78$ meters (to 2 decimal places)
Commentary: The height of the cocoa tree is approximately 16.78 meters. This method is practical for estimating heights of tall objects without direct measurement.
Construction and Engineering (e.g., Building a house, Road Construction): Trigonometry is used to determine the pitch of a roof (angle of the roof slope) to ensure proper water runoff. Builders use tangent to calculate the height of a gable end given the width of the house and the desired roof angle. Engineers use sine and cosine to calculate the gradient of roads and ramps, ensuring they are not too steep for vehicles or wheelchairs (e.g., designing ramps for disabled access in public buildings in Abuja or Lagos). Architects use trigonometric ratios to calculate heights of structures, lengths of support beams, and overall structural stability. Surveying and Land Management (e.g., Farmland Measurement, Property Boundaries): Surveyors in Nigeria use trigonometry to measure distances and angles for mapping land, establishing property boundaries, and planning infrastructure projects. For instance, to determine the area of an irregularly shaped piece of farmland in a rural community, surveyors might break it into triangles and use trigonometric principles to find unknown sides or angles, which then help in calculating area. It is used in creating contour maps and determining heights of hills or depths of valleys, which is vital for agricultural planning and water management. Navigation (e.g., River Transport, Aviation): Pilots and ship captains use trigonometry to calculate distances, directions, and altitudes/depths. For example, a boat captain navigating the Niger Delta waterways can use trigonometric ratios to determine the distance to a landmark or the width of a river, given an angle and a known distance from a point. Air traffic controllers use trigonometry to track aircraft positions, calculate their altitudes, and predict flight paths, ensuring safe air travel over Nigerian airspace.