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
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