The document RD Sharma Solutions -Ex-18.1, Basic Geometrical Tools, Class 6, Maths Class 6 Notes | EduRev is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.

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**1.) Construct the following angles using set- squares:**

**Answer:**

**(i) 45 ^{o}**

Place 45^{o} set- square.

Draw two rays AB and AC along the edges from the vertex from the vertex of 45^{o} angle of the set- square.

The angle so formed is a 45^{o} angle.

âˆ BAC=45^{o}

**(ii) 90 ^{o}**

Place = 90^{o} set â€“square as shown in the figure.

Draw two rays BC and BA along the edges from the vertex of 90^{o} angle.

The angle so formed is 90^{o} angle.

âˆ ABC=90^{o}

**(iii) 60 ^{o}**

Place 30^{o} set â€“square as shown in the figure.

Draw the rays BA and BC along the edges from the vertex of 60^{o}

The angle so formed is 60^{o}

âˆ ABC=60^{o}

**(iv) 105 ^{o}**

Place 30^{o} set â€“square and make an angle 60^{o} by drawing the rays BA and BC as shown in figure.

Now place the vertex of 45^{o} of the set â€“square on the ray BA as shown in figure and draw the ray BD.

The angle so formed is 105^{o}

Therefore, âˆ DBC=105^{o}

**(v ) 75 ^{o}**

Place 45^{o} set â€“square and make an angle of 45^{o} by drawing the rays BD and BC as shown in the figure.

Now place the vertex of 30^{o} of the set- square on the ray BD as shown in the figure and draw the ray BA.

The angle so formed is 75^{o}.

Therefore, âˆ ABC=75^{o}

(Line BD is hidden)

**(vi) 150 ^{o}**

Place the vertex of 45^{o} of the set â€“ square and make angle of 90^{o} by drawing the rays BD and BC as shown in the figure

Now, place the vertex of 30^{o} of the set â€“square on the ray BS as shown in the figure and draw the ray BA

The angle so formed is 150^{o}.

Therefore, âˆ ABC=150^{o}

**2.) Given a line BC and a point A on it, construct a ray AD using set â€“ squares so that âˆ DAC is**

**(i) 30 ^{o}**

**(ii) 150 ^{o}**

**Answer:**

(i) Draw a line BC and take a point A on it. Place 30^{o} set â€“square on the line BC such that its vertex of 30^{o} angle lies on point A and one edge coincides with the ray AB as shown in figure

Draw the ray AD.

Thus âˆ DAC is the required angle of 30^{o}

(ii) Draw a line BC and take a point A on it. Place 30^{o} set â€“square on the line BC such that its vertex of 30^{o} angle lies on point A and one edge coincides with the ray AB as shown in the figure.

Draw the ray AD.

Therefore, âˆ DAB=30^{o}

We know that angle on one side of the straight line will always add to 180^{o}

Therefore, âˆ DAB+âˆ DAC=180^{o}

Therefore, âˆ DAC=150^{o}