In Fig. 5.28, PQ ∥ RS. If ∠1 = (2a + b)° and ∠6 = (3a − b)°, then the measure of ∠2 in terms of b is

(a) (2 + b)°
(b) (3 − b)°
(c) (108 − b)°
(d) (180 − b)°
Given ∠1 = 2a+b and ∠6 = 3a−b are alternate interior angles (equal).
So 2a+b = 3a−b ⇒ a=2b.
Now ∠2 = ∠1 = 2a+b = 4b+b=5b.
But since lines parallel, ∠2 + ∠6 = 180 ⇒ 5b+(3a−b)=180. Substituting a=2b ⇒ 5b+(6b−b)=180 ⇒ 10b=180 ⇒ b=18. ∠2=5b=90. Matches form (108−b).