Question

A man in a boat rowing away from a lighthouse 150m high,takes 2 minutes to change the angle of elevation of the top of the lighthouse from 45 to 30 degree. Find the speed of the boat.
plzzzzzz....fast..........

Answer 1

Answer:

0.915 m/s or 54.9 m/min.

Step-by-step explanation:

AB ➝ Lighthouse.

The angle of elevation from point C ➝ ∠ACB = 45°

The angle of elevation from point D ➝ ∠ACB = 30°

Time taken to travel from point C to D ➝ 2 minutes.

[Check the diagram attached for markings]

In ΔABC.

∠B = 90°

∴ We can say that:

$\Longrightarrow \sf tan45^\circ = \dfrac{Opposite \ side}{Adjacent \ Side}$

The side opposite to 45° is AB and the side adjacent to it is CB.

And, we know that tan45° = 1.

$\Longrightarrow \sf 1 = \dfrac{AB}{CB}$

$\Longrightarrow \sf CB = AB$

AB = 150m, therefore:

$\Longrightarrow \sf CB = 150m.$

In ΔABD.

∠B = 90°

∴ We can say that:

$\Longrightarrow \sf tan30^\circ = \dfrac{Opposite \ side}{Adjacent \ Side}$

The side opposite to 30° is AB and the side adjacent to it is DB.

And, we know that tan30° = 1/√3.

$\Longrightarrow \sf \dfrac{1}{\sqrt{3}} = \dfrac{AB}{DB}$

DB can be written as DC + CB.

$\Longrightarrow \sf \dfrac{1}{\sqrt{3}} = \dfrac{150}{DC + CB}$

$\Longrightarrow \sf \dfrac{1}{\sqrt{3}} = \dfrac{150}{DC + 150}$

$\Longrightarrow \sf DC + 150 = 150\sqrt{3}$

$\Longrightarrow \sf DC = 150\sqrt{3} - 150$

$\Longrightarrow \sf DC = 150\Big[\sqrt{3} - 1\Big]$

$\Longrightarrow \sf DC = 150\Big[1.732 - 1\Big]$

$\Longrightarrow \sf DC = 150\Big[0.732\Big]$

$\Longrightarrow \sf DC = 109.8m$

Now that we have the distance covered, let's find the speed of the boat.

$\Longrightarrow \sf Speed \ of \ the \ boat = \dfrac{Distance \ travelled}{Time \ taken}$

$\Longrightarrow \sf Speed \ of \ the \ boat = \dfrac{109.8}{2}$

$\Longrightarrow \sf Speed \ of \ the \ boat = 54.9 \ m/min$

From converting m/min to m/s we get:

$\Longrightarrow \sf Speed \ of \ the \ boat \ in \ m/s = 54.9 \times \dfrac{1}{60} \ m/s$

$\Longrightarrow \sf Speed \ of \ the \ boat \ in \ m/s = \bf{0.915 \ m/s}$

Therefore, the speed of the boat is 0.915 m/s.

Answer 2

0.915m/sec

Step-by-step explanation:

In ∆ABC

tan 45° = AB

CB

tan 45° = 150

y

1 = 150

y

y = 150 ..........(i)

In ∆ABD

tan 30° = AB

DC + CB

tan 30° = 150

x + y

Putting value of y

1 = 150

√3 x + 150

x + 150 = 150√3

x = 150 ( √3 - 1 )

Speed = Distance

Time

speed = DC = 150 ( √3 - 1 ) = 54.9

2 min 2

converting minutes into seconds:

54.9 X 1 = 0.915m/sec

60