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Geometry and Measurement - Angles on a Line, at a Point, and in Triangles

Grade 7IB

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

🔑Concepts

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Angles on a straight line add up to 180∘180^{\circ}. This is known as a supplementary angle relationship. If a straight line is divided into multiple angles at a point, their sum is always equal to a straight angle (180∘180^{\circ}).

A straight line with a ray dividing it into two angles a and b summing to 180 degrees.
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Angles around a point (a full turn) always sum to 360∘360^{\circ}. Regardless of how many rays originate from a single vertex, the total rotation equals 360∘360^{\circ}.

Three rays meeting at a central point forming angles x, y, and z that sum to 360 degrees.
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The sum of the interior angles of any triangle is always 180∘180^{\circ}. This property holds true for scalene, isosceles, equilateral, and right-angled triangles.

A triangle with interior angles labeled A, B, and C.
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The exterior angle of a triangle is equal to the sum of the two opposite interior angles. This is a direct consequence of the triangle angle sum and straight line angle properties.

A triangle with one side extended to show an exterior angle equal to the sum of opposite interior angles a and b.

📐Formulae

Sum of angles on a line=180∘\text{Sum of angles on a line} = 180^{\circ}

Sum of angles at a point=360∘\text{Sum of angles at a point} = 360^{\circ}

∠A+∠B+∠C=180∘ (Triangle interior sum)\angle A + \angle B + \angle C = 180^{\circ} \text{ (Triangle interior sum)}

Exterior angle (d)=∠a+∠b (where a,b are opposite interior angles)\text{Exterior angle } (d) = \angle a + \angle b \text{ (where } a, b \text{ are opposite interior angles)}

Angle in an equilateral triangle=180∘3=60∘\text{Angle in an equilateral triangle} = \frac{180^{\circ}}{3} = 60^{\circ}

💡Examples

Problem 1:

Calculate the value of xx if three angles on a straight line are given as xx, 42∘42^{\circ}, and 88∘88^{\circ}.

Solution:

  1. Use the property that angles on a straight line sum to 180∘180^{\circ}: x+42∘+88∘=180∘x + 42^{\circ} + 88^{\circ} = 180^{\circ} 2. Add the known values together: x+130∘=180∘x + 130^{\circ} = 180^{\circ} 3. Subtract 130∘130^{\circ} from both sides to solve for xx: x=180∘−130∘x = 180^{\circ} - 130^{\circ} 4. The result is: x=50∘x = 50^{\circ}

Explanation:

This problem is solved by identifying that the angles are supplementary because they sit on a straight line, meaning their total must be 180∘180^{\circ}.

Problem 2:

An isosceles triangle has a vertex angle (the angle between the two equal sides) of 40∘40^{\circ}. What is the size of each base angle?

Solution:

  1. Let each base angle be represented by bb. Since it is an isosceles triangle, both base angles are equal. 2. Set up the triangle sum equation: 40∘+b+b=180∘40^{\circ} + b + b = 180^{\circ} 3. Simplify the equation: 40∘+2b=180∘40^{\circ} + 2b = 180^{\circ} 4. Subtract 40∘40^{\circ} from both sides: 2b=140∘2b = 140^{\circ} 5. Divide by 22 to find the value of one base angle: b=140∘2=70∘b = \frac{140^{\circ}}{2} = 70^{\circ}

Explanation:

In an isosceles triangle, we subtract the vertex angle from 180∘180^{\circ} and divide the remainder by 22 because the two remaining angles are equal.

Problem 3:

In the given diagram, four angles meet at a point. Three of the angles are 110∘110^{\circ}, 85∘85^{\circ}, and 75∘75^{\circ}. Find the value of the fourth angle yy.

Four angles meeting at a point with three labeled values and one variable y.

Solution:

110∘+85∘+75∘+y=360∘110^{\circ} + 85^{\circ} + 75^{\circ} + y = 360^{\circ} 270∘+y=360∘270^{\circ} + y = 360^{\circ} y=360∘−270∘y = 360^{\circ} - 270^{\circ} y=90∘y = 90^{\circ}

Explanation:

Since the angles are at a point, they must sum to 360∘360^{\circ}. By adding the known angles and subtracting from 360∘360^{\circ}, we find the missing angle.

Problem 4:

Find the value of kk in the triangle shown, where the interior angles are 2k2k, 3k3k, and 50∘50^{\circ}.

A triangle with angles labeled 2k, 3k, and 50 degrees.

Solution:

2k+3k+50∘=180∘2k + 3k + 50^{\circ} = 180^{\circ} 5k+50∘=180∘5k + 50^{\circ} = 180^{\circ} 5k=130∘5k = 130^{\circ} k=130∘5k = \frac{130^{\circ}}{5} k=26∘k = 26^{\circ}

Explanation:

The sum of interior angles in a triangle is 180∘180^{\circ}. We set up an equation summing the three expressions to 180180 and solve for the unknown variable kk.

Angles on a Line, at a Point, and in Triangles Grade 7 Notes & Examples