**New Learning Composite Mathematics Class 6 SK Gupta Anubhuti Gangal Basic Constructions Chapter 13 Solution**

**Process:**

**(a) **At first I draw a ray BC. Write B as centre and any convenient radius I draw an arc, cutting BC at Q. With Q as center and same radius I draw another arc to cut the first arc at P. Now I join BP. Thus ∠PBQ = 60^{0}.

Then with compass I bisect ∠PBQ, BO is the angel bisector.

Thus ∠ABO = ∠OBC = 30^{0}

**(b) **At first I draw a ray OR. Then I construct an angel ∠PQR = 90^{0}. Then with compass I bisect ∠PQR, QS is the angel bisector.

Thus ∠PQS = ∠SQR = 45^{0}

**Process:**

**(c) **At first I draw a line AC. With a vertex B or AC. I construct ∠FBC = 150^{0}.

Then with compass I bisect ∠FBC, BD is the angel bisector.

Thus ∠FBD = ∠DBC = 75^{0}

**(d) **At first I draw a line AC. With a vertex B on AC, I construct ∠FBC = 135^{0}

Then with compass I bisect ∠FBC, BD is the angel bisector.

Thus ∠FBD = ∠DBC = 67 1/2^{0 }

**Process:**

**(e) **I draw a line AC, with point B on the AC as vertex, construct ∠CBD = 90^{0}. Then ∠ABD = 90^{0}

Then, ∠CBE = 90^{0} + 45^{0} = 135^{0}

**(f) **I draw a line AC. With a vertex B on , I construct ∠CBD = 120^{0}.

Then, I bisect ∠ABD, BE is the angel bisector

Thus, ∠EBC = 120^{0} + 30^{0} = 150^{0}

**(g) **I draw a line AC with B as centre and any convenient radius I draw an arc to cut the BC at X with X as centre and the radius, I draw another arc to cut the first arc at Y. with Y as centre and the same radius, I draw another arc to cut the another arc at Z. Then I draw the ray BD through B and Z. Thus ∠CBD = 120^{0}