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Practical Geometry - Constructing Angles of specific measures (60°, 30°, 120°, 90°, 45°)

Grade 6CBSE

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

🔑Concepts

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The 60∘60^{\circ} Angle Construction: This is the fundamental building block of compass-based geometry. To visualize this, draw a ray OAOA. With OO as the vertex, draw a wide arc cutting OAOA at a point. Keeping the compass width exactly the same, place the pointer on that point and draw another arc intersecting the first one. The line through this intersection from OO creates a 60∘60^{\circ} angle.

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The 120∘120^{\circ} Angle Construction: This is essentially two 60∘60^{\circ} constructions joined together. Visually, after marking the first 60∘60^{\circ} intersection on your main arc, you move the compass pointer to that new intersection and mark a second point further along the same arc. Since each step covers 60∘60^{\circ}, the second point is 120∘120^{\circ} from the starting ray.

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Angle Bisector Principle: This technique is used to divide any angle into two equal parts. For example, to get 30∘30^{\circ}, you bisect 60∘60^{\circ}. Visually, you place the compass on the two points where the angle's arms meet an arc and draw two crossing arcs in the 'middle' space. The line from the vertex through this cross is the bisector.

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The 90∘90^{\circ} (Right Angle) Construction: A 90∘90^{\circ} angle is found exactly halfway between 60∘60^{\circ} and 120∘120^{\circ}. To visualize this, after marking the 60∘60^{\circ} and 120∘120^{\circ} points on your initial arc, draw two intersecting arcs using those points as centers. Connecting the vertex to this intersection creates a perpendicular line (90∘90^{\circ}).

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Constructing 45∘45^{\circ} and 30∘30^{\circ}: These are 'derived' angles created by halving larger angles. 45∘45^{\circ} is the bisector of a 90∘90^{\circ} angle, and 30∘30^{\circ} is the bisector of a 60∘60^{\circ} angle. Visually, the resulting ray will always split the 'mouth' of the original angle into two perfectly symmetrical sections.

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Radius Consistency: In all these constructions, the most important visual rule is keeping the compass radius fixed while marking the 60∘60^{\circ} and 120∘120^{\circ} points. Changing the width mid-construction will result in an inaccurate angle measure.

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Verification with Protractor: While practical geometry focuses on construction using a ruler and compass, the final step should always be visual verification using a protractor to ensure the angle measures exactly the intended degrees.

📐Formulae

Angle Bisector=12×Original Angle\text{Angle Bisector} = \frac{1}{2} \times \text{Original Angle}

30∘=60∘230^{\circ} = \frac{60^{\circ}}{2}

45∘=90∘245^{\circ} = \frac{90^{\circ}}{2}

90∘=60∘+120∘−60∘290^{\circ} = 60^{\circ} + \frac{120^{\circ} - 60^{\circ}}{2}

Straight Angle=180∘\text{Straight Angle} = 180^{\circ}

💡Examples

Problem 1:

Construct an angle of 60∘60^{\circ} at the endpoint OO of a ray OAOA.

Solution:

  1. Draw a ray OAOA. 2. With OO as center and any convenient radius, draw an arc intersecting OAOA at point PP. 3. With PP as center and the same radius as before, draw an arc intersecting the first arc at point QQ. 4. Draw a ray OBOB passing through QQ. 5. The measure of ∠BOA\angle BOA is 60∘60^{\circ}.

Explanation:

This construction uses the property of an equilateral triangle. Since the radius (distance from OO to PP, OO to QQ, and PP to QQ) is kept constant, △OPQ\triangle OPQ would be equilateral, meaning all angles are 60∘60^{\circ}.

Problem 2:

Construct an angle of 90∘90^{\circ} and use it to construct a 45∘45^{\circ} angle.

Solution:

  1. Draw a line LL and mark a point OO on it. 2. Draw a semi-circle arc with center OO cutting the line at XX and YY. 3. From XX, mark the 60∘60^{\circ} arc and then the 120∘120^{\circ} arc. 4. Bisect the space between 60∘60^{\circ} and 120∘120^{\circ} to find the 90∘90^{\circ} ray, OCOC. 5. To get 45∘45^{\circ}, place the compass at the point where the arc hits the horizontal line and where it hits the 90∘90^{\circ} ray OCOC. 6. Draw two arcs that intersect at point DD and join ODOD. ∠DOA=45∘\angle DOA = 45^{\circ}.

Explanation:

We first create a perpendicular (90∘90^{\circ}) by finding the midpoint between 60∘60^{\circ} and 120∘120^{\circ}. Then, we apply the angle bisector method to the 90∘90^{\circ} angle to halve it into 45∘45^{\circ}.