Given: Radius OA = 75 cm, AB is tangent at A, so ∠OAB = 90°, and ∠ABO = 30°.
(a) Length of AB:
In △OAB, tan∠ABO = OA/AB
tan 30° = 75/AB
AB = 75/tan 30° = 75 × √3 = 75√3 cm
(b) Length of OB:
sin∠ABO = OA/OB
sin 30° = 75/OB
1/2 = 75/OB
OB = 150 cm
(c) Length of AP:
Since PQ ∥ OA and Q is on the circle, PQ is also perpendicular to AB (as OA ⊥ AB).
In △PQB, ∠QPB = 90°, ∠PBQ = 30°.
Now, BQ is a secant/chord and using properties: Since PQ ∥ OA, △BPQ ~ △BOA.
BP/BO = BQ/BA (by similar triangles... let BP = x)
x/150 = BQ/75√3
Also, BQ = BO – OQ... Let's use similarity ratio:
BP/OB = PQ/OA, and BP/OB = AB – AP / OB
Since △BPQ ~ △BOA:
BP/BO = PB/150
Using ratio: BP/150 = PQ/75 → need another relation.
By similarity: BP/OB = BQ/BA → BP = OB × (BQ/BA)
Since PQ ∥ OA, P divides AB and Q divides OB-line proportionally.
AP = AB − BP
By similar triangles (△BPQ ~ △BOA):
BP/OB = BP/150 ⟹ ratio = BP/150
BQ/BA = BP/BO ⟹ BQ = OB·(BP/OB) ...
Using correct similarity: BP/BO = BQ/BA = PQ/OA = k
Let BP/150 = k, so BP = 150k and BQ = 75√3·k.
Since Q lies on circle, OQ = 75. In △OBQ: not straightforward.
Using direct ratio: Since PQ ∥ OA, by Basic Proportionality:
BP/BO = BQ/BA (not applicable here directly).
In △BOA, line PQ ∥ OA cuts BO at Q and BA at P:
→ BQ/BO = BP/BA
BQ = BO − OQ = 150 − 75 = 75 cm
∴ BP/BA = BQ/BO = 75/150 = 1/2
BP = (1/2) × 75√3 = 75√3/2 cm
AP = AB − BP = 75√3 − 75√3/2 = 75√3/2 cm
$$\boxed{AP = \dfrac{75\sqrt{3}}{2} \text{ cm}}$$
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