So, the data is inadequate.

Then, ∠ACB = 60° and AC = 4.6 m.

Draw BE ⊥ CD.

Then, CE = AB = 1.6 m,

      BE = AC = 20√3 m.

∴ CD = CE + DE = (1.6 + 20) m = 21.6 m.

Then, ∠APB = 30° and AB = 100 m.

Let ∠ACB = Θ.

∴ Θ = 30°.

Then, AB = 100 m, ∠ACB = 30° and ∠ADB = 45°.

Consider the diagram is shown above where PR represents the ladder and RQ represent the wall.
Cos 60° = PQ / PR
? 1 / 2 = 12.4 / PR
? PR = 2 × 12.4 = 24.8 m

Let us draw a figure below as per given question.
Let A be the position of a person on the bank of a river and OP the tree on the opposite bank and ?OAP = 60°. When the person retires to the position B, then AB = 40 meter and ?OBP = 30°
Let us assume OA(Breadth of the river) = x meter and height of tree OP = h meter
In ?OAP, Use the trigonometry formula
Tan60° = P/B = Perpendicular distance / Base distance
? Tan60° = OP / OA
? OP = OA Tan60°
Put the value of OP and OA, We will get
? h = x?3 ..............(1)
Now in the triangle ?OBP
Tan30° = OP / OB
? OP = OB Tan30°
? OP = (x + 40)/?3
? h = (x + 40)/?3 ...................(2)
From Equation (1) and (2), We will get
? (x + 40)/?3 = x?3
? (x + 40) = x?3 X ?3
? (x + 40) = 3x
? 3x - x = 40
? x = 20 m

Let us draw the figure from the given question.
Let AB = h meter be the height of the tower; C and D are the two points on the ground such that BC = 60 m; ?ACB = 45° and ?ADB = 30°
Now from right triangle ABC,
tan 45° = h/60
? 1 = h/60
? h = 60 m;
Again from right triangle ABD;
tan 30° = h/(x + 60)
? 1/?3 = 60/(x + 60)
? x + 60 = 60?3
? x = 60(1.73 - 1) = 43.8 meter
Hence, speed of boat = 43.8/5 m/s = 43.8/5 x 18/5 = 31.5 km/hr.

Let us draw a figure from given question.
Let, AB = 100 m be the height of a tower, P is a point on the ground such that ∠APB = 30°
From right triangle ABP,
BP = 100 cot 30°
⇒ BP = 100 √3
⇒ BP = 100 X 1.73
⇒ BP = 173 meter