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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 is the foundational construction in practical geometry. By drawing an arc with a compass and then, without changing the radius, drawing a second arc from the point where the first arc intersects the base line, you create an equilateral triangle's vertex, which is exactly 60∘60^{\circ}.

Construction of a 60 degree angle using a compass and straightedge showing the intersecting arcs.
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An angle of 120∘120^{\circ} is constructed by marking two consecutive 60∘60^{\circ} arcs on a large primary arc. Since 120∘=2×60∘120^{\circ} = 2 \times 60^{\circ}, the second intersection point from the base line represents 120∘120^{\circ} from the starting ray.

Construction of a 120 degree angle showing two 60 degree steps on a semi-circle arc.
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A 90∘90^{\circ} angle (perpendicular) can be constructed by bisecting the angle between 60∘60^{\circ} and 120∘120^{\circ}. Because 90∘90^{\circ} is exactly halfway between 60∘60^{\circ} and 120∘120^{\circ}, we use the intersection points of these two arcs to find the vertical line.

Construction of a 90 degree angle by bisecting the region between 60 and 120 degrees.
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Sub-angles like 30∘30^{\circ} and 45∘45^{\circ} are created through angle bisector techniques. 30∘30^{\circ} is the bisector of a 60∘60^{\circ} angle, and 45∘45^{\circ} is the bisector of a 90∘90^{\circ} angle.

📐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}.

Problem 3:

Construct an angle of 150∘150^{\circ} using only a compass and ruler.

Geometry construction of a 150 degree angle.

Solution:

  1. Draw a line segment PQPQ and mark point OO on it.
  2. Construct a 120∘120^{\circ} angle and a 180∘180^{\circ} angle (straight line) on the same arc.
  3. Bisect the angle between 120∘120^{\circ} and 180∘180^{\circ} because 150∘=120∘+180∘−120∘2150^{\circ} = 120^{\circ} + \frac{180^{\circ} - 120^{\circ}}{2}.
  4. The resulting ray OROR makes an angle ∠ROQ=150∘\angle ROQ = 150^{\circ}.

Explanation:

To construct 150∘150^{\circ}, we identify that it lies exactly halfway between 120∘120^{\circ} and 180∘180^{\circ}. By bisecting that 60∘60^{\circ} gap, we add 30∘30^{\circ} to 120∘120^{\circ}.

Problem 4:

Construct an angle of 75∘75^{\circ} using a compass and a ruler.

Construction of a 75 degree angle by bisecting the 60 to 90 degree arc segment.

Solution:

  1. Draw a ray OAOA.
  2. Construct a 90∘90^{\circ} angle and a 60∘60^{\circ} angle on the same primary arc.
  3. The space between 60∘60^{\circ} and 90∘90^{\circ} is 30∘30^{\circ}.
  4. Bisect this 30∘30^{\circ} interval to get 15∘15^{\circ}.
  5. 60∘+15∘=75∘60^{\circ} + 15^{\circ} = 75^{\circ}. The ray OBOB passing through this bisector forms ∠BOA=75∘\angle BOA = 75^{\circ}.

Explanation:

A 75∘75^{\circ} angle is found by bisecting the region between the 60∘60^{\circ} and 90∘90^{\circ} construction marks.