krit.club logo

Geometry - Angle Properties (Lines, Triangles, Quadrilaterals)

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

When two parallel lines are cut by a transversal, corresponding angles are equal. They form an 'F' shape in the diagram.

Diagram showing corresponding angles a and b formed by parallel lines and a transversal.
•

Alternate angles are equal. These are often called 'Z' angles and are located between parallel lines on opposite sides of the transversal.

Diagram showing alternate angles x and y between parallel lines.
•

In an isosceles triangle, the angles opposite the equal sides are equal.

Isosceles triangle with base angles marked as equal.
•

Vertically opposite angles are equal whenever two straight lines intersect.

Two intersecting lines showing equal vertically opposite angles 1 and 2.

📐Formulae

Angles on a line: a+b+c=180∘a + b + c = 180^{\circ}

Sum of angles in a triangle: ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^{\circ}

Exterior angle of triangle: ext∠=int∠opp1+int∠opp2ext\angle = int\angle_{opp1} + int\angle_{opp2}

Sum of angles in a quadrilateral: ∠A+∠B+∠C+∠D=360∘\angle A + \angle B + \angle C + \angle D = 360^{\circ}

Sum of interior angles of an nn-sided polygon: (n−2)×180∘(n - 2) \times 180^{\circ}

Co-interior angles: x+y=180∘x + y = 180^{\circ} (where lines are parallel)

💡Examples

Problem 1:

In triangle ABC, angle A is 40∘40^{\circ} and angle B is 85∘85^{\circ}. Calculate the exterior angle at vertex C.

Solution:

125∘125^{\circ}

Explanation:

Using the exterior angle theorem, the exterior angle is equal to the sum of the two opposite interior angles. Therefore, Exterior ∠C=∠A+∠B=40∘+85∘=125∘\angle C = \angle A + \angle B = 40^{\circ} + 85^{\circ} = 125^{\circ}.

Problem 2:

Two parallel lines are intersected by a transversal. If one of the co-interior angles is 72∘72^{\circ}, find the value of the other co-interior angle.

Solution:

108∘108^{\circ}

Explanation:

Co-interior (allied) angles between parallel lines are supplementary, meaning they add up to 180∘180^{\circ}. Calculation: 180∘−72∘=108∘180^{\circ} - 72^{\circ} = 108^{\circ}.

Problem 3:

A quadrilateral has three angles measuring 110∘110^{\circ}, 80∘80^{\circ}, and 75∘75^{\circ}. Find the size of the fourth angle.

Solution:

95∘95^{\circ}

Explanation:

The sum of angles in a quadrilateral is always 360∘360^{\circ}. Sum of known angles: 110+80+75=265∘110 + 80 + 75 = 265^{\circ}. Fourth angle: 360∘−265∘=95∘360^{\circ} - 265^{\circ} = 95^{\circ}.

Problem 4:

In the diagram below, L1L_1 and L2L_2 are parallel. Find the value of xx.

Parallel lines L1 and L2 with a transversal. An angle of 130 degrees is shown at L1 and angle x is shown at L2.

Solution:

  1. Identify the relationship: The angle 130∘130^{\circ} and the angle adjacent to xx on the straight line are corresponding angles, or we can use alternate angles.
  2. The angle alternate to 130∘130^{\circ} is 130∘130^{\circ}.
  3. Since xx and the alternate angle lie on a straight line: x+130∘=180∘x + 130^{\circ} = 180^{\circ}
  4. Subtract 130∘130^{\circ} from both sides: x=180∘−130∘x = 180^{\circ} - 130^{\circ} x=50∘x = 50^{\circ}

Explanation:

We use the property that alternate angles are equal and then apply the property that angles on a straight line sum to 180∘180^{\circ}.

Problem 5:

Find the value of yy in the following kite. One vertex angle is 100∘100^{\circ} and the two equal opposite angles are 110∘110^{\circ} each.

A kite with angles 100, 110, 110 and y degrees.

Solution:

  1. A kite is a quadrilateral, so the sum of interior angles is 360∘360^{\circ}.
  2. Let the unknown angle be yy.
  3. Sum the known angles: 110∘+110∘+100∘=320∘110^{\circ} + 110^{\circ} + 100^{\circ} = 320^{\circ}
  4. Subtract from 360∘360^{\circ}: y=360∘−320∘y = 360^{\circ} - 320^{\circ} y=40∘y = 40^{\circ}

Explanation:

The sum of angles in any quadrilateral is 360∘360^{\circ}. By subtracting the three known angles of the kite from 360∘360^{\circ}, we find the missing angle.