In the given figure, two straight lines PQ and RS intersect each other at O. If ∠POT = 75°, find the values of a + b + c.

Here, 4b + 75° + b = 180°   [A straight angle]
5b = 180° – 75° = 105°
b = $\frac{{105}^o}{5}$ = 21°
∴ a = 4b = 4 × 21° = 84°   [Vertically opposite angles]
Again, 2c + a = 180°   [A linear pair]
⇒ 2c + 84° = 180°
⇒ 2c = 96°
⇒ c = $\frac{{96}^o}{2}$ = 48°
Hence, the values of a, b and c are:

a = 84°, b = 21°, and c = 48°

So, a + b + c = 84° + 21° + 48° = 153^o

Therefore, option C is the correct answer.