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Trigonometric Ratios Of Some Special Angles

NIOS · Class 10 · Maths

Flashcards for Trigonometric Ratios Of Some Special Angles — NIOS Class 10 Maths. Quick Q&A cards covering key concepts, definitions, and formulas.

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A labeled diagram showing a right-angled isosceles triangle (angles 45°, 45°, 90°) with sides 'a' and hypotenuse 'a√2', used to derive sine, cosine, and tangent for 45 degrees.
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Card 1Trigonometric ratios for 45 degrees

Solve: In a right triangle, if one acute angle is 45° and one side adjacent to it is a units, find the hypotenuse.

Answer

Step 1: For 45°, the two legs are equal, so the other leg is also a units. Step 2: Use Pythagoras theorem: hypotenuse^2 = a^2 + a^2 = 2a^2. Step 3: Hypotenuse = sqrt(2)a. Answer: The hypotenuse is sqr

Card 2Special angle values

Find sin 45°, cos 45°, tan 45°, cosec 45°, sec 45°, and cot 45°.

Answer

Step 1: In a 45°-45°-90° triangle, both legs are equal and the hypotenuse is sqrt(2) times one leg. Step 2: sin 45° = 1/sqrt(2). Step 3: cos 45° = 1/sqrt(2). Step 4: tan 45° = 1. Step 5: cosec 45° = s

Card 3Trigonometric ratios for 30 degrees

Solve: In the 30° construction, if PM = a units, find OP and OM.

Answer

Step 1: In the 30° construction, the triangle formed becomes equilateral with side 2a. Step 2: Therefore OP = 2a units. Step 3: Use Pythagoras theorem in the right triangle: OP^2 = PM^2 + OM^2. Step 4

Card 4Special angle values

Find sin 30°, cos 30°, tan 30°, cosec 30°, sec 30°, and cot 30°.

Answer

Step 1: Use the 30° special triangle values. Step 2: sin 30° = 1/2. Step 3: cos 30° = sqrt(3)/2. Step 4: tan 30° = 1/sqrt(3). Step 5: cosec 30° = 2. Step 6: sec 30° = 2/sqrt(3). Step 7: cot 30° = sqrt

Card 5Trigonometric ratios for 60 degrees

Solve: In the 60° construction, if OM = a units, find PM and OP.

Answer

Step 1: In the 60° construction, the triangle formed becomes equilateral with side 2a. Step 2: Therefore OP = 2a units. Step 3: Use Pythagoras theorem in the right triangle: OP^2 = OM^2 + PM^2. Step 4

Card 6Special angle values

Find sin 60°, cos 60°, tan 60°, cosec 60°, sec 60°, and cot 60°.

Answer

Step 1: Use the 60° special triangle values. Step 2: sin 60° = sqrt(3)/2. Step 3: cos 60° = 1/2. Step 4: tan 60° = sqrt(3). Step 5: cosec 60° = 2/sqrt(3). Step 6: sec 60° = 2. Step 7: cot 60° = 1/sqrt

Card 7Special angle evaluation

Evaluate: tan^2 60° - sin^2 30°

Answer

Step 1: tan 60° = sqrt(3), so tan^2 60° = 3. Step 2: sin 30° = 1/2, so sin^2 30° = 1/4. Step 3: Subtract: 3 - 1/4 = 11/4. Answer: 11/4.

Card 8Special angle evaluation

Evaluate: cot^2 30° sec^2 45° + cosec^2 45° cos 60°

Answer

Step 1: cot 30° = sqrt(3), so cot^2 30° = 3. Step 2: sec 45° = sqrt(2), so sec^2 45° = 2. Step 3: cosec 45° = sqrt(2), so cosec^2 45° = 2. Step 4: cos 60° = 1/2. Step 5: Compute: (3)(2) + (2)(1/2) = 6

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What are the important topics in Trigonometric Ratios Of Some Special Angles for NIOS Class 10 Maths?
Key topics in Trigonometric Ratios Of Some Special Angles include Trigonometric Ratios of Special Angles - Complete Overview, Complete Overview of Special Angles Chapter, Special Angles Trigonometric Ratios Overview. These are the concepts NIOS Class 10 examiners draw on most — study them first, then practise related questions.
How to score full marks in Trigonometric Ratios Of Some Special Angles — NIOS Class 10 Maths?
Understand the core concepts first, then work through the 36 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
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