Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Trigonometric Ratios
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Trigonometric Ratios
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Subject: General Mathematics
Class: Senior Secondary 1
Term: 1st Term
Week: 8
Theme: Geometry
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Teacher Activity: Briefly review Pythagoras' Theorem and its application in right-angled triangles. Draw various right-angled triangles on the board and ask students to identify the hypotenuse, opposite, and adjacent sides relative to a given acute angle. Introduce the concept of ratios of sides in similar right-angled triangles being constant for a given angle. This leads to the idea of trigonometric ratios.
Student Activity: Respond to questions on Pythagoras' Theorem. Identify and label sides of right-angled triangles. Participate in a brief discussion on the constant nature of side ratios in similar triangles.
A right-angled triangle is a triangle with one angle measuring 90°. The sides of a right-angled triangle are named relative to a reference acute angle (θ).
Hypotenuse (H): The side opposite the right angle, always the longest side.
Opposite (O): The side directly opposite the reference angle (θ).
Adjacent (A): The side next to the reference angle (θ), which is not the hypotenuse. The three primary trigonometric ratios are: Sine (sin θ): The ratio of the length of the opposite side to the length of the hypotenuse. $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{O}{H}$ (SOH) Cosine (cos θ): The ratio of the length of the adjacent side to the length of the hypotenuse. $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{A}{H}$ (CAH) Tangent (tan θ): The ratio of the length of the opposite side to the length of the adjacent side. $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{O}{A}$ (TOA)
Mnemonic: "SOH CAH TOA" helps remember these ratios. Worked Example 2.1.1 (Calculating a side length): A ladder 5m long leans against a vertical wall, making an angle of 60° with the ground. How high up the wall does the ladder reach?
Solution: Sketch the diagram: Draw a right-angled triangle. Hypotenuse = 5m (ladder length) Angle θ = 60° (angle with the ground) Opposite side = height (h) (what we need to find) Adjacent side = distance from wall to base of ladder (not needed here)
Identify the knowns and unknowns: H = 5m θ = 60° O = h (unknown)
Choose the appropriate ratio: We have O and H, so use Sine. $\sin \theta = \frac{O}{H}$ $\sin 60^\circ = \frac{h}{5}$ Solve for h: $h = 5 \times \sin 60^\circ$ (Students will learn $\sin 60^\circ = \frac{\sqrt{3}}{2}$ in the next section) $h = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}$ m Using $\sqrt{3} \approx 1.732$, $h \approx \frac{5 \times 1.732}{2} = \frac{8.66}{2} = 4.33$ m. The ladder reaches approximately 4.33m high up the wall. Worked Example 2.1.2 (Calculating an angle): A surveyor in Kaduna measures the distance from point A to the base of a telecom mast (point B) as 80m. The height of the mast (BC) is 60m. Calculate the angle of elevation of the top of the mast from point
A. Solution: Sketch the diagram: Draw a right-angled triangle ABC, with the right angle at B. Adjacent side (AB) = 80m Opposite side (BC) = 60m Angle of elevation (θ) at A (unknown)
Identify knowns and unknowns: A = 80m O = 60m θ (unknown)
Choose the appropriate ratio: We have O and A, so use Tangent. $\tan \theta = \frac{O}{A}$ $\tan \theta = \frac{60}{80} = \frac{3}{4} = 0.75$ Solve for θ: $\theta = \tan^{-1}(0.75)$ Using a calculator (if permitted, otherwise this would be stated as the answer), $\theta \approx 36.87^\circ$. The angle of elevation is approximately 36.87°. Consider an isosceles right-angled triangle with two equal sides of length 1 unit. By Pythagoras' theorem, Hypotenuse = $\sqrt{1^2 + 1^2} = \sqrt{2}$. The acute angles are both 45°. For $\theta = 45^\circ$: Opposite = 1 Adjacent = 1 Hypotenuse = $\sqrt{2}$ $\sin 45^\circ = \frac{O}{H} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $\cos 45^\circ = \frac{A}{H} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $\tan 45^\circ = \frac{O}{A} = \frac{1}{1} = 1$ Consider an equilateral triangle with side length 2 units. All angles are 60°. Draw an altitude (height) from one vertex to the midpoint of the opposite side. This altitude bisects the angle at the vertex (creating 30°) and the base (creating a segment of length 1). This forms a right-angled triangle. Hypotenuse = 2 One leg = 1 By Pythagoras' theorem, the altitude (other leg) = $\sqrt{2^2 - 1^2} = \sqrt{4-1} = \sqrt{3}$. For $\theta = 60^\circ$: Opposite = $\sqrt{3}$ Adjacent = 1 Hypotenuse = 2 $\sin 60^\circ = \frac{O}{H} = \frac{\sqrt{3}}{2}$ $\cos 60^\circ = \frac{A}{H} = \frac{1}{2}$ $\tan 60^\circ = \frac{O}{A} = \frac{\sqrt{3}}{1} = \sqrt{3}$ For $\theta = 30^\circ$: Opposite = 1 Adjacent = $\sqrt{3}$ Hypotenuse = 2 $\sin 30^\circ = \frac{O}{H} = \frac{1}{2}$ $\cos 30^\circ = \frac{A}{H} = \frac{\sqrt{3}}{2}$ $\tan 30^\circ = \frac{O}{A} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ Summary Table of Special Angle Ratios: | Angle $(\theta)$ | $\sin \theta$ | $\cos \theta$ | $\tan \theta$ | | :--------------- | :------------------ | :------------------ | :------------------ | | 30° | $1/2$ | $\sqrt{3}/2$ | $1/\sqrt{3}$ or $\sqrt{3}/3$ | | 45° | $1/\sqrt{2}$ or $\sqrt{2}/2$ | $1/\sqrt{2}$ or $\sqrt{2}/2$ | $1$ | | 60° | $\sqrt{3}/2$ | $1/2$ | $\sqrt{3}$ | Worked Example 2.2.1 (Applying special angles): A street vendor in Lagos sights the top of a local landmark (e.g., National Arts Theatre) at an angle of elevation of 30°. If the vendor is 100m away from the base of the landmark, what is its height? (Assume the vendor's eye level is negligible).
Solution: Sketch the diagram: Right-angled triangle. Adjacent = 100m Angle θ = 30° Opposite = height (h)
Choose the ratio: We have A and want O, so use Tangent. $\tan \theta = \frac{O}{A}$ $\tan 30^\circ = \frac{h}{100}$ Substitute special angle value: $\frac{1}{\sqrt{3}} = \frac{h}{100}$ Solve for h: $h = \frac{100}{\sqrt{3}} = \frac{100\sqrt{3}}{3}$ m The height of the landmark is $\frac{100\sqrt{3}}{3}$ meters. (No calculator needed). These angles are important because their trigonometric ratios can be derived geometrically and expressed as exact values, often used without calculators.
Land Surveying and Construction in Rural Areas: In many parts of Nigeria, traditional land surveying might involve using simple tools. Trigonometry allows for calculating inaccessible distances. For instance, determining the width of a local river (e.g., River Benue or a stream in a village) without crossing it, by measuring an angle of depression from a known height on one bank and using the tangent ratio. This can inform decisions about bridge building or pipe laying. Determining Heights of Local Landmarks or Structures: Students can apply trigonometric ratios to calculate the height of common structures in their communities, such as school buildings, telecom masts, water tanks, or even prominent trees. For example, a student could measure their distance from a water tank and the angle of elevation to its top using a simple clinometer (made from a protractor, string, and weight), then use the tangent ratio to find its height. This makes mathematics tangible and relevant to their environment. Safety and Accessibility in Public Infrastructure: The concepts of angles of elevation are critical in designing ramps for accessibility (e.g., for wheelchairs at a public hospital or market). Students can understand how a shallow angle (smaller angle of elevation) makes a ramp longer but easier to use, while a steeper angle makes it shorter but harder. This connects trigonometry to inclusive design and community well-being.