Two ships leave a port at the same time. One goes 24km/hr in the direction N45°E and the other travels 32km/hr in the direction S75°E. Find the distance between the ships at the end of 3 hours.

 

 

1.jpg

Image.jpg

 

Text version of the answer is

Let Ox be the position of the first ship and dy be the position  of the second ship. At the end of 3 hours.

Speed of ship 1a : 34 km/h

Distance travelled in

    1 hour : 24 km

Distance travelled in

   3 hours  = 0X = 24 × 3

      OX = 72 km

Speed of ship  : 32 km/hr

Distance traelled in 1  hour : 32 km

Distance travelled in 3 hours = 32 × 3  = 96 km 

 

In ∆ OXY, by the using cosine formula,

 (xy) 2 = (ox) 2 + (Oy) 2 − 2 (OX) (OY) cos ∠XOY ….. (1)

 

We know ∠NOS = 180°

    ∠XON + ∠XOY + ∠YOS  = 180°

             45° + ∠XOY + 75°  = 180°

                       120° + ∠xoy = 180°    

                                  ∠xoy = 60°

 

Hence, Put OX = 72 km, OY = 96 km and

∠XOY = 60°  in equation (1)     

 

(Xy 2 ) = (72) 2 + (96) 2 − 2 (72) (96) cos 60°

 (Xy 2 ) = (72) 2 + (96) 2 − 2 (72) (96) (1/2)

(Xy 2 ) = 5154 + 9216 − 6912

(Xy 2 ) = 7488

Xy = √7488

     = √(26×32×13)

     = √((23×3)2)  √(×13)

     = 24√13 km

     = 24 × 3.60 km

     = 86.4 km

 

Rough

  2 | 7488

  2 | 3744

  2 | 1872

  2 | 936

  2 | 468

  2 | 234

  3 | 117

  3 | 39

 13 | 13

  2 | 1

= √7488 = 2 × 2 × 2 × 3 × √13

 


About the author
Teachoo's image
Teachoo