Question

A man is watching a boat speeding away from the top of a tower. The boat makes an angle of depression of 60° with the man’s eye when at a distance of 200 m from the tower. After 10 seconds, the angle of depression becomes 45°. What is the approximate speed of the boat (in km / hr), assuming that it is sailing in still water? ( √3 = 1.732)

Answer 1

Answer:

Let AB be the tower and C and D be the point of observation from boat

Given AC=60 m And ∠ADB=300 and ∠ACB=450

Let height of tower is h m

In ΔBAC

Thentan450=ACAB

⇒1=ACAB⇒AB=AC=60m

In ΔBAD

Then tan30o=ADAB

⇒31=ADAB

⇒AD=3AB=603m$$

∴CD=AD−AC=603−60=60(3−1)

Then speed=560(3−1) m/s

And approximate speed =100012×0.73×3600≈32 km/h



Answer 2

Answer:

• Speed of the boat = 52.70 km / hr

Step-by-step explanation:

Let,

• AB be the tower.
• C and D be the position of the boat.

From the diagram,

• ∠XAC = 60° = ∠ACB and
• ∠XAD = 45° = ∠ADB,
• BC = 200 m

In the right angled ΔABC,

⟶ tan 60° = AB / BC

⟶ $\sf\sqrt{3}$ = AB / 200

⟶ AB = 200$\sf\sqrt{3}$ --------(1)

In the right angled ΔABD,

⟶ tan 45° = AB / BD

⟶ 1 = 200$\sf\sqrt{3}$ / BD

⟶ BD = 200$\sf\sqrt{3}$

⟶ CD = BD - BC

⟶ CD = 200$\sf\sqrt{3}$ - 200($\sf\sqrt{3}$ - 1)

⟶ 146.4

It is given that the Distance CD is covered in 10 seconds.

This is, the distance of 146.4 m is covered in 10 seconds.

Speed of the boat,

⟶ Speed = Distance / Time

⟶ 146.4 / 10

⟶ 14.64 m / s

⟶ 14.64 × (3600 / 1000) km / hr

⟶ 52.70 km / hr

• ∴ The speed of the boat = 52.70 km/hr