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.

 

Answer

1.jpg

2.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