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How do you use angle properties of triangles, polygons, parallel lines and circles to find missing angles?

Using the angle properties of triangles, quadrilaterals and polygons, angles in parallel lines, and the symmetry and angle properties of the circle to calculate missing angles.

A focused answer to the SQA National 5 Mathematics shape and angle content, covering angle sums of triangles and polygons, angles in parallel lines, the angle in a semicircle, the tangent-radius right angle, and using the symmetry of the circle to find missing angles.

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  1. What this dot point is asking
  2. Angles in triangles and polygons
  3. Angles in parallel lines
  4. Angles in the circle
  5. Examples in context
  6. Try this

What this dot point is asking

The SQA wants you to use angle facts to calculate missing angles in triangles, quadrilaterals and polygons, in parallel-line diagrams, and in circles, justifying each step with the property you used. Clear reasoning matters as much as the final number.

Angles in triangles and polygons

The angle sum of a triangle is always 180∘180^\circ, and of a quadrilateral 360∘360^\circ. For any polygon, split it into triangles from one vertex.

Angles in parallel lines

When a straight line crosses a pair of parallel lines, several equal and supplementary angles appear.

To solve a parallel-line problem, name the angle you can find, state the property, and chain the facts until you reach the required angle.

Angles in the circle

The circle has its own angle properties, several arising from its symmetry. Two radii to the ends of a chord form an isosceles triangle, which lets you find base angles.

Examples in context

Angle properties underpin construction and design. A roof truss is a triangle whose angles must sum to 180∘180^\circ, so a designer can find the third angle from the other two. Tiling patterns rely on regular polygons whose interior angles fit exactly around a point (360∘360^\circ). Engineers use the tangent-radius right angle when a belt leaves a pulley, since the belt is tangent to the wheel.

Try this

Q1. Find the missing angle in a triangle with angles 40∘40^\circ and 75∘75^\circ. [1 mark]

  • Cue. 180βˆ’40βˆ’75=65∘180 - 40 - 75 = 65^\circ.

Q2. Find each interior angle of a regular octagon. [2 marks]

  • Cue. (8βˆ’2)Γ—1808=135∘\dfrac{(8 - 2) \times 180}{8} = 135^\circ.

Q3. ABAB is a diameter and CC is on the circle. Angle ABC=60∘ABC = 60^\circ. Find angle BACBAC. [2 marks]

  • Cue. Angle ACB=90∘ACB = 90^\circ, so angle BAC=30∘BAC = 30^\circ.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA National 5 20193 marksA regular polygon has 99 sides. Calculate the size of each interior angle.
Show worked answer β†’

The interior angles of an nn-sided polygon sum to (nβˆ’2)Γ—180∘(n - 2) \times 180^\circ, so for n=9n = 9 the sum is 7Γ—180∘=1260∘7 \times 180^\circ = 1260^\circ (2 marks). A regular polygon has equal angles, so each is 12609=140∘\dfrac{1260}{9} = 140^\circ (1 mark). Markers reward the angle sum and dividing by the number of sides.

SQA National 5 20223 marksABAB is a diameter of a circle, and CC is a point on the circumference. Angle CAB=35∘CAB = 35^\circ. Calculate angle ABCABC.
Show worked answer β†’

The angle in a semicircle is a right angle, so angle ACB=90∘ACB = 90^\circ (1 mark). The angles of triangle ABCABC sum to 180∘180^\circ: 35∘+90∘+angleΒ ABC=180∘35^\circ + 90^\circ + \text{angle } ABC = 180^\circ (1 mark). So angle ABC=180βˆ˜βˆ’125∘=55∘ABC = 180^\circ - 125^\circ = 55^\circ (1 mark). Markers reward using the semicircle right angle and the triangle angle sum.

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