Question

a man in a Boat moving away from a Lighthouse 100m high takes 2 minutes to change the angle of elevation of the top of the Lighthouse from 60 degree to 30 degree find the speed of the boat in metre per minute and use root 3 is equal to 1.732​

Answer 1

Answer:

The speed of the boat while changing the elevation is  57.745 meters per minutes .

Step-by-step explanation:

Given as :

The height of the lighthouse = h = 100 meters

The angle of elevations are $\Theta _1$ = 60°  ,  $\Theta _2$ = 30°

The time taken for elevation change = t = 2 minutes

Let The speed of the boat = s meters per min

According to question

Tan angle = $\dfrac{\textrm perpendicular}{\textrm base}$

In Δ OAC

Tan 60°  = $\dfrac{OC}{OA}$

or, 1.732 = $\dfrac{h}{x}$

or, 1.732 = $\dfrac{100}{x}$

∴   x = 57.73 meters

So The distance between OA = x = 57.73 meters

Again

In Δ OBC

Tan 30° = $\dfrac{OC}{OB}$

Or, 0.5773 = $\dfrac{h}{y}$

Or, 0.5773 = $\dfrac{100}{y}$

∴ y = 173.22

So, The change in position of boat = OB - OA

i.e The change in position of boat = y - x

Or, The change in position of boat = 173.22 m - 57.73 m

∴ The change in position of boat = 115.49 meters

Now

The distance cover by boat during change in elevation = d = 115.49 m

And The time taken to change the elevation = t = 2 min

So, Speed of boat = $\dfrac{Distance}{time}$

Or, s = $\dfrac{d}{t}$

i.e s = $\dfrac{115.49}{2}$

Or, speed = 57.745 meters per minutes

So, The speed of the boat = s =  57.745 meters per minutes

Hence, The speed of the boat while changing the elevation is  57.745 meters per minutes . Answer