## RS Aggarwal Class 6 Solutions Chapter 13 Angles and Their Measurement Ex 13C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 13 Angles and Their Measurement Ex 13C.

**Other Exercises**

- RS Aggarwal Solutions Class 6 Chapter 13 Angles and Their Measurement Ex 13A
- RS Aggarwal Solutions Class 6 Chapter 13 Angles and Their Measurement Ex 13B
- RS Aggarwal Solutions Class 6 Chapter 13 Angles and Their Measurement Ex 13C
- RS Aggarwal Solutions Class 6 Chapter 13 Angles and Their Measurement Ex 13D

**Question 1.**

**Solution:**

(i) Place the protractor in such a way that its centre is exactly at the vertex O of the given angle AOB and the base line lies along the arm OA. Read off the mark through which the arm OB passes, starting from 0° on the side A.

We find that ∠AOB = 45°.

(ii) The given angle is ∠PQR. Place the protractor in such a way that its centre is exactly on the vertex Q of the given angle and the base line lies along the arm QR.Read off the mark through which the arm QP passes, starting from 0° on the side of R.

We find that ∠PQR = 67°

(in) The given angle is ∠DEF. Place the protractor in such a way that its centre is exactly on the vertex E of the given angle and the base line lies along the arm ED. Read off the mark through which the arm EF passes, starting from 0° on the side of D.

We find that ∠DEF = 130°

(iv) The given angle is ∠LMN. Place the protractor in such a way that its centre is exactly on the vertex M of the given angle and the base line lies along the arm ML. Read off the mark through which the arm MN passes, starting from 0° on the side of L.

We find that ∠LMN = 50°

(v) The given angle is ∠RST. Place the protractor in such a way that its centre is exactly on the vertex S of the given angle and the base line lies along the arm SR. Read off the mark through which the arm ST passes, starting from 0° on the side of R.

We find that the ∠RST = 130°.

(vi) The given angle is ∠GHI. Place the protractor in such a way that its centre is exactly on the vertex H of the given angle and the base line lies along the arm HI. Read off the mark through which the arm HG passes, starting from 0° on the side of I.

We find that ∠GHI = 70°

**Question 2.**

**Solution:**

(i) Draw a ray OA. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OA. Starting from 0° on the side of A, look for the 25° mark on the protractor. Mark a point B at this 25° mark. Remove the protractor and draw the ray OB. Then ∠AOB is the required angle.

(ii) Draw a ray OA. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OA. Starting from 0° on the side of A, look for the 72° mark on the protractor. Mark a point B at this 72° mark. Remove the protractor and draw the ray OB. Then,

∠AOB is the required angle of measure 72°.

(iii) Draw a ray OA. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OA. Starting from 0° on the side of A, look for the 90° mark on the protractor. Mark a point B at this 90° mark. Remove the protractor and draw the ray OB. Then ∠AOB is the required angle whose measure is 90°.

(iv) Draw a ray OA. Place the protractor in such a way that its centre exactly lies at O and the base line lies along OA. Starting from 0° on the side of A, look for the 117° mark on the protractor. Mark a point B at this 117° mark. Remove the protractor and draw the ray OB. Then ∠AOB is the required angle whose measure is 117°.

(v) Draw a ray OP. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OP. Starting from 0° on the side of P, look for the 165° mark on the protractor. Mark a point Q at this 165° mark. Remove the protractor and draw the ray OQ. Then, ∠POQ is the required angle whose measure is 165°.

(vi) Draw a ray OP. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OP. Starting from 0° on the side of P, look for the 23° mark on the protractor. Mark a point Q at this 23° mark. Remove the protractor and draw the ray OQ. Then ∠POQ is the required angle whose measure is 23°.

(vii) Draw a ray OA. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OA. Starting from 0° on the side of A,Took for the 180° mark on the protractor. Mark a point B on this 180° mark. Remove the protractor and draw the ray OB. Then ∠AOB is the required angle whose measure is 180°.

(viii) Draw a ray Rs. Place the protractor in such away that its centre lies exactly at R and the base line lies along RS. Starting from 0° on the side of S, look for the 48° mark on the protractor. Mark a point T at this 48° mark. Remove the protractor and draw the ray RT. Then, ∠SRT is the required angle whose measure is 48°.

**Question 3.**

**Solution:**

On measuring the given angle ABC with the help of a protractor, it is 50°

Now, place the protractor on EF in such a way that its centre lies on E exactly and base with the line EF.

Now read off the mark through with the arm ED passes at 50°.

Join DE,

Then ∠DEF is equal to 50° i.e. equal to ∠ABC.

**Question 4.**

**Solution:**

Steps of construction :

(i) Draw a line segment AB = 6 cm.

(ii) Take a point C on AB such that AC = 4 cm.

(iii) Place protractor with its centre at C and base along CB.

(iv) Mark a point D against 90°.

(v) Remove the protractor and join DC. Then DC ⊥ AB. Ans.

Hope given RS Aggarwal Solutions Class 6 Chapter 13 Angles and Their Measurement Ex 13C are helpful to complete your math homework.

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