(i) 45°
Construction :
Step 1: Drawn a line and marked a point A on it.
Step 2: With A as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. ∠BAD = 120°.
Step 6: With G as center and any convenient radius drawn an arc in the interior of ∠GAB
Step 7: With the same radius and B as center drawn an arc to cut the arc at F.
Step 8: Joined AF. ∠BAF = 45°
(ii) 150°
Construction :
Since 150° = 60° + 60° + 30°; we construct as follows
Step 1: Drawn a line and marked a point A on it.
Step 2: With ‘A’ as center, drawn a full arc of convenient radius to the line at a point B and at E the other end.
Step 3: With the same radius and B as center, drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center drawn an arc to cut the already drawn arc at D.
Step 5: With D as center, drawn an arc of convenient radius in the interior of ∠DAE
Step 6: With E as center and with the same radius drawn an arc to cut the previous arc at F.
Step 7: Joined AF, ∠FAB = 150°.
(iii) 135°
Construction :
Step 1: Drawn a line and marked a point A on it.
Step 2: With ‘A’ as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc at D.
Step 5: With C and D as centers drawn arcs of convenient (same) radius in the interior of ∠CAD. Marked the point of intersection as E.
Step 6: Joined AE, through G. ∠BAE = 90°.
Step 7: Drawn angle bisector to ∠GAH through F.
Now ∠BAF = 135°.
