Maths Mathematics

MathematicsUnit 116 min read

Properties of Triangles: Angles, Sides, Theorems & Trigonometry

Unit 11 of Mathematics covers the fundamental properties of triangles, including angle and side relationships, similarity and congruence theorems, and trigonometric ratios. This note explains key concepts with step-by-step examples, visual proofs, and NEB-style practice questions.

TAKEAWAYS:

  • Angles in a triangle always sum to 180° and follow specific relationships (e.g., exterior angle = sum of two opposite interior angles).
  • Similar triangles have equal angles and proportional sides, solved using AA (Angle-Angle), SAS (Side-Angle-Side), or SSS (Side-Side-Side) criteria.
  • Congruent triangles are identical in shape and size, verified by SSS, SAS, ASA, AAS, or RHS rules.
  • Trigonometric ratios (sine, cosine, tangent) relate angles to side lengths in right triangles and are used to solve real-world problems.
  • Pythagoras’ theorem applies only to right triangles: .
  • Area and height formulas for triangles include and Heron’s formula for scalene triangles.

1. Basic Properties of Triangles

Triangles are three-sided polygons with three vertices and three angles. Their properties depend on their sides and angles.

[object Object]ABC6 cm8 cm10 cm60°53°67°
Scalene triangle with sides 6 cm, 8 cm, 10 cm and angles 60°, 53°, 67°

Types of Triangles by Sides

Types of Triangles by Angles

Key Angle Properties

  1. Sum of interior angles = . figure {"type":"right-triangle","sides":{"hypotenuse":"c","opposite":"a","adjacent":"b","angle":"θ"},"labels":{"sin":"a/c","cos":"b/c","tan":"a/b"},"caption":"Trigonometric ratios in a right triangle"}
- **sin θ** = \(\frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{a}{c}\)
- **cos θ** = \(\frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{b}{c}\)
- **tan θ** = \(\frac{\text{Opposite}}{\text{Adjacent}} = \frac{a}{b}\)

#### **Worked Example: Finding Side Lengths**
**Problem**: In △ABC (right-angled at B), ∠A = 30°, and AC = 10 cm. Find BC and AB.
**Solution**:
1. Draw the triangle:
   ```figure
   {"type":"right-triangle","sides":{"hypotenuse":"AC = 10 cm","opposite":"BC","adjacent":"AB","angle":"30°"},"caption":"Right triangle with ∠A = 30°"}
  1. Use sin 30°: [ \sin 30° = \frac{BC}{AC} \implies \frac{1}{2} = \frac{BC}{10} \implies BC = 5 \text{ cm} ]
  2. Use cos 30°: [ \cos 30° = \frac{AB}{AC} \implies \frac{\sqrt{3}}{2} = \frac{AB}{10} \implies AB = 5\sqrt{3} \text{ cm} ]
  3. Answer: BC = 5 cm, AB = (5\sqrt{3}) cm.

5. Pythagoras’ Theorem

In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: [ a^2 + b^2 = c^2 ] Worked Example: Problem: A ladder leans against a wall at 6 m height. If the base is 8 m from the wall, find the ladder’s length. Solution:

  1. Draw the triangle:
    ABC
    Ladder problem
  2. Apply Pythagoras’ theorem: [ 6^2 + 8^2 = c^2 \implies 36 + 64 = c^2 \implies c^2 = 100 \implies c = 10 \text{ m} ]
  3. Answer: The ladder is 10 m long.
ABC
Pythagorean theorem: a² + b² = c² (visual proof with squares)

6. Area of Triangles

ABC
Triangle with base 12 cm and height 9 cm (Area = 1/2 × base × height)

Formula 1: Base × Height / 2

[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ] Example: Base = 8 cm, height = 5 cm → Area = (\frac{1}{2} \times 8 \times 5 = 20 \text{ cm}^2).

Formula 2: Heron’s Formula (for scalene triangles)

  1. Find the semi-perimeter (s = \frac{a + b + c}{2}).
  2. Area = (\sqrt{s(s - a)(s - b)(s - c)}). Example: Sides = 5 cm, 6 cm, 7 cm.
  3. (s = \frac{5 + 6 + 7}{2} = 9).
  4. Area = (\sqrt{9(9-5)(9-6)(9-7)} = \sqrt{9 \times 4 \times 3 \times 2} = \sqrt{216} = 6\sqrt{6} \text{ cm}^2).

Exam Tip

  1. Memorize angle sum properties, similarity/congruence criteria, and trigonometric ratios.
  2. Draw diagrams for every problem—visualizing helps!
  3. Label sides/angles clearly in proofs (e.g., "Given: AB = DE, ∠B = ∠E").
  4. Practice NEB-style questions:
    • Short answer: "State the similarity criterion used if two triangles have angles 50°, 60°, 70°." Answer: AA (Angle-Angle) criterion.
    • Proof: "Prove △ABC ≅ △DEF given AB = DE, BC = EF, and ∠B = ∠E." Answer: Use SSS (all sides equal).
    • Calculation: "Find the height of a triangle with base 12 cm and area 36 cm²." Answer: (36 = \frac{1}{2} \times 12 \times h \implies h = 6 \text{ cm}).
flowchart TD
    A["Memorize Key Properties"] --> B["Draw Diagrams"]
    B --> C["Label Clearly"]
    C --> D["Practice NEB Questions"]
    D --> E["Check Answers"]
Step-by-step exam preparation strategy for triangles

NEB-Style Practice Questions

  1. In △PQR, ∠P = 40°, ∠Q = 70°. Find ∠R and an exterior angle at Q.
  2. Two triangles have sides in ratio 3:5 and included angles 45°. Are they similar? Why?
  3. A 10 m pole casts a 6 m shadow. Find the angle of elevation of the sun.
  4. Prove △ABC ≅ △DEF if AB = DE, ∠B = ∠E, and BC = EF.
  5. Find the area of a triangle with sides 9 cm, 12 cm, and 15 cm using Heron’s formula.

Key Formulas Summary

Concept Formula
Sum of angles in △
Exterior angle
Similarity (AA) Two angles equal
Congruence (SAS) Two sides and included angle equal
Trigonometric ratios , ,
Pythagoras’ theorem
Area (base × height)
Heron’s formula

Based on the NEB +2 Science syllabus for Mathematics (Maths), unit 11.

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