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Properties Of Triangles

Telangana Open School (TOSS) · Class 12 · Mathematics

Flashcards for Properties Of Triangles — Telangana Open School (TOSS) Class 12 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.

47 questions25 flashcards5 concepts

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25 Flashcards
Card 1Cosine Rule

In triangle ABC, if a = 7, b = 5, and ∠C = 60°, find side c using the cosine rule.

Answer

Using cosine rule: c² = a² + b² - 2ab cos C Step 1: a = 7, b = 5, ∠C = 60°, cos 60° = 0.5 Step 2: c² = 7² + 5² - 2×7×5×0.5 = 49 + 25 - 35 = 39 Step 3: c = √39 ≈ 6.24 cm Answer: c ≈ 6.24 cm

Card 2Cosine Rule

If in triangle ABC, a = 6, b = 8, and c = 10, find ∠C using the cosine formula.

Answer

Using cosine formula: cos C = (a² + b² - c²) / (2ab) Step 1: a = 6, b = 8, c = 10 Step 2: cos C = (36 + 64 - 100) / (2×6×8) = (100 - 100)/96 = 0/96 = 0 Step 3: cos C = 0 ⇒ ∠C = 90° Answer: ∠C = 90°

Card 3Sine Rule

In triangle ABC, ∠A = 45°, ∠B = 60°, and side a = 10 cm. Find side b using the sine rule.

Answer

Using sine rule: a/sin A = b/sin B Step 1: a = 10, ∠A = 45°, ∠B = 60° Step 2: 10 / sin 45° = b / sin 60° Step 3: sin 45° = √2/2 ≈ 0.707, sin 60° = √3/2 ≈ 0.866 Step 4: b = 10 × (0.866 / 0.707) ≈ 10 ×

Card 4Cosine Rule

The sides of a triangle are 5 cm, 7 cm, and 8 cm. Find the largest angle using the cosine rule.

Answer

Largest angle is opposite largest side (8 cm), so find ∠C. Using: cos C = (a² + b² - c²)/(2ab) Step 1: a = 5, b = 7, c = 8 Step 2: cos C = (25 + 49 - 64)/(2×5×7) = (74 - 64)/70 = 10/70 = 1/7 ≈ 0.1429

Card 5Cosine Rule

In triangle ABC, a = 4, b = 5, c = 6. Find cos A, cos B, and cos C.

Answer

Using cosine formulas: For cos A: (b² + c² - a²)/(2bc) = (25 + 36 - 16)/(2×5×6) = 45/60 = 0.75 For cos B: (a² + c² - b²)/(2ac) = (16 + 36 - 25)/(2×4×6) = 27/48 = 0.5625 For cos C: (a² + b² - c²)/(2ab)

Card 6Cosine Rule

If in triangle ABC, a cos A = b cos B and a ≠ b, prove it is right-angled.

Answer

Given: a cos A = b cos B Using cos A = (b² + c² - a²)/(2bc), cos B = (c² + a² - b²)/(2ca) So: a×(b² + c² - a²)/(2bc) = b×(c² + a² - b²)/(2ca) Multiply both sides by 2c: a(b² + c² - a²)/b = b(c² + a² -

Card 7Cosine Rule

In triangle ABC, a = 3, b = 4, c = 5. Find cos A, cos B, and cos C.

Answer

Using cosine formulas: cos A = (b² + c² - a²)/(2bc) = (16 + 25 - 9)/(2×4×5) = 32/40 = 0.8 cos B = (a² + c² - b²)/(2ac) = (9 + 25 - 16)/(2×3×5) = 18/30 = 0.6 cos C = (a² + b² - c²)/(2ab) = (9 + 16 - 25

Card 8Sine Rule

When do you use the sine rule?

Answer

Use the sine rule when: 1. Two angles and one side are known (AAS or ASA) 2. Two sides and a non-included angle are known (SSA) Example: In triangle ABC, ∠A = 30°, ∠B = 45°, a = 6 cm. Find b. Solution

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Frequently Asked Questions

What are the important topics in Properties Of Triangles for Telangana Open School (TOSS) Class 12 Mathematics?
Properties Of Triangles covers several key topics that are frequently asked in Telangana Open School (TOSS) Class 12 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Properties Of Triangles — Telangana Open School (TOSS) Class 12 Mathematics?
Understand the core concepts first, then work through the 47 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
How many flashcards are available for Properties Of Triangles?
There are 25 flashcards for Properties Of Triangles covering key definitions, formulas, and concepts. Use them daily for 10–15 minutes for best results.

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