\

1.3.1) \ \

\

Destination distance = 600 km in west direction. \ \

\

Destination time t_d= 55 minutes \ \

\

t_d= 55/60 = 0.917 hr \ \

\

Wind  blowing speed v_w=   75 m/s in north-west direction. \ \

\

v_w=   75 * 3.6 = 270 km/hr \ \

\

∠COA = Relative angle \ \

\

Speed of flight = distance / time = 600 / 0.917 = 654 km/hr \ \

\

OA = 645 km/hr \ \

\

OB = 270 km/hr \ \

\

Resultant speed DE² = OA² + OB² - 2 OA OB cos45 \ \

\

DE² = 645² + 270² - 2 (270)(645) cos45 \ \

\

DE² = 645² + 270² - 2 (270)(645) cos45 \ \

\

DE = 492.58 km/hr \ \

\

For parallelogram

\

(diagonal1)² + (diagonal2)² = (side1 + side2)² \ \

\

(DE)² + (OC)² = (OE + OD)² \ \

\

(OC)² = ( 645 + 270 )² - 492.58² \ \

\

OC = 771.5 km/hr. \ \

\

From OAC right angled triangle \ \

\

cosθ = OA/OC \ \

\

cosθ = 645/771.5 \ \

\

cosθ = 0.836 \ \

\

θ = 33.3^o \ \

\

Solution :

\

Direction of the velocity of plane relative to the velocity of ground is 33.3° upside inclined to west

\