Introduction to Trigonometry Ex 8.1: Ratios, Examples & Solutions

Understanding Trigonometric Ratios in a Right-Angled Triangle

Everything in Exercise 8.1 revolves around a right-angled triangle. Let's consider a triangle ABC, right-angled at B.

The three sides have special names:

1. Hypotenuse: This is the longest side, and it's always opposite the right angle (90°). In our triangle, AC is the hypotenuse.
2. Perpendicular (or Opposite Side): This side is opposite the angle we are considering.
3. Base (or Adjacent Side): This side is next to the angle we are considering (and is not the hypotenuse).

Crucial Point: The labels 'Perpendicular' and 'Base' change depending on which angle (other than the 90° angle) you are focusing on!

• For Angle A: The side opposite is BC (so, Perpendicular = BC). The side adjacent is AB (so, Base = AB).
• For Angle C: The side opposite is AB (so, Perpendicular = AB). The side adjacent is BC (so, Base = BC).

The Hypotenuse (AC) remains the same for both angles. Understanding this switch is the key to mastering this exercise.

The Six Trigonometric Ratios

sine (sin)

The ratio of the length of the Perpendicular to the Hypotenuse. (sin θ = P/H)

cosine (cos)

The ratio of the length of the Base to the Hypotenuse. (cos θ = B/H)

tangent (tan)

The ratio of the length of the Perpendicular to the Base. (tan θ = P/B)

cosecant (cosec)

The reciprocal of sine. (cosec θ = 1/sin θ = H/P)

secant (sec)

The reciprocal of cosine. (sec θ = 1/cos θ = H/B)

cotangent (cot)

The reciprocal of tangent. (cot θ = 1/tan θ = B/P)

Worked Example: Finding All Ratios from a Given Ratio

1. Problem Statement — Given that tan A = 4/3, find the other trigonometric ratios of the angle A.
2. Step 1: Understand the Given Ratio and Draw a Triangle — We know that tan A = Perpendicular / Base. So, we can say P = 4k and B = 3k, where k is a positive number. Let's draw a right-angled triangle ABC, right-angled at B. For angle A, the Perpendicular is BC = 4k and the Base is AB = 3k.
3. Step 2: Find the Third Side using Pythagoras' Theorem — Pythagoras' Theorem states that (Hypotenuse)² = (Perpendicular)² + (Base)². AC² = BC² + AB² AC² = (4k)² + (3k)² AC² = 16k² + 9k² AC² = 25k² AC = √(25k²) = 5k. So, the Hypotenuse is 5k.
4. Step 3: Calculate the Remaining Ratios — Now we have all three sides: P = 4k, B = 3k, and H = 5k. sin A = P/H = 4k/5k = 4/5 cos A = B/H = 3k/5k = 3/5 cosec A = 1/sin A = 5/4 sec A = 1/cos A = 5/3 * cot A = 1/tan A = 3/4

Key Points & Common Errors in Ex 8.1

1. Confusing Perpendicular and Base: This is the most common mistake. Always identify the angle you are working with first. The side opposite that angle is the perpendicular.

2. Forgetting Pythagoras' Theorem: If you are given one ratio (like tan A = 4/3), you only know two sides proportionally. You MUST use Pythagoras' theorem to find the third side before you can calculate the other ratios.

3. Notation Error: Do not write sin A as 'sin × A'. The term 'sin' has no meaning by itself; it is a function of an angle. 'sin A' is a single, indivisible term representing the sine of angle A.

4. Forgetting the 'k': When a ratio like sin A = 3/5 is given, it means the sides are in the ratio 3:5. The actual lengths could be 3cm and 5cm, or 6cm and 10cm, etc. It is mathematically correct to write the sides as 3k and 5k. Although 'k' cancels out in the final ratios, using it shows a proper understanding of the concept.

Practice Questions with Solutions

• Q: If sin A = 3/5, find cos A and tan A. A: Step 1: In a right-angled triangle, sin A = Perpendicular/Hypotenuse = 3/5. Let Perpendicular (P) = 3k and Hypotenuse (H) = 5k. Step 2: Using Pythagoras theorem, Base (B)^2 = H^2 - P^2 = (5k)^2 - (3k)^2 = 25k^2 - 9k^2 = 16k^2. So, B = 4k. Step 3: cos A = Base/Hypotenuse = 4k/5k = 4/5. tan A = Perpendicular/Base = 3k/4k = 3/4. Final answer: cos A = 4/5, tan A = 3/4.
• Q: If cot θ = 7/8, evaluate (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1 - cos θ). A: Step 1: Given cot θ = 7/8. In a right triangle, cot θ = Base/Perpendicular. So, let Base (B) = 7k and Perpendicular (P) = 8k. Step 2: Using Pythagoras theorem, Hypotenuse (H)^2 = P^2 + B^2 = (8k)^2 + (7k)^2 = 64k^2 + 49k^2 = 113k^2. So, H = sqrt(113)k. Step 3: sin θ = P/H = 8k/(sqrt(113)k) = 8/sqrt(113). cos θ = B/H = 7k/(sqrt(113)k) = 7/sqrt(113). Step 4: The expression is (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1 - cos θ) = (1 - sin^2 θ) / (1 - cos^2 θ). Step 5: Substitute the values: (1 - (8/sqrt(113))^2) / (1 - (7/sqrt(113))^2) = (1 - 64/113) / (1 - 49/113) = ( (113-64)/113 ) / ( (113-49)/113 ) = (49/113) / (64/113) = 49/64. Final answer: 49/64.
• Q: In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine sin A and cos C. A: Step 1: Given AB = 24 cm and BC = 7 cm. These are the two legs of the right-angled triangle. Step 2: Using Pythagoras theorem, Hypotenuse AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625. So, AC = sqrt(625) = 25 cm. Step 3: For angle A: Perpendicular = BC = 7 cm, Hypotenuse = AC = 25 cm. So, sin A = BC/AC = 7/25. Step 4: For angle C: Base = BC = 7 cm, Hypotenuse = AC = 25 cm. So, cos C = BC/AC = 7/25. Final answer: sin A = 7/25, cos C = 7/25.
• Q: Given 15 cot A = 8, find sec A. A: Step 1: From the given equation, 15 cot A = 8, we can write cot A = 8/15. Step 2: In a right-angled triangle, cot A = Base/Perpendicular. So, let Base (B) = 8k and Perpendicular (P) = 15k. Step 3: Using Pythagoras theorem, Hypotenuse (H)^2 = P^2 + B^2 = (15k)^2 + (8k)^2 = 225k^2 + 64k^2 = 289k^2. So, H = sqrt(289)k = 17k. Step 4: sec A = Hypotenuse/Base = 17k/8k = 17/8. Final answer: sec A = 17/8.

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