Answer:
(a)123 km/hr
(b)39 degrees
Step-by-step explanation:
Plane X with an average speed of 50km/hr travels for 2 hours from T (Kano Airport) to point U in the diagram.
Distance = Speed X Time
Therefore: Distance from T to U =50km/hr X 2 hr =100 km
It moves from Point U at 9.00 am and arrives at the airstrip A by 11.30am.
Distance, UA=50km/hr X 2.5 hr =125 km
Using alternate angles in the diagram:
[tex]\angle U=110^\circ[/tex]
(a)First, we calculate the distance traveled, TA by plane Y.
Using Cosine rule
[tex]u^2=t^2+a^2-2ta\cos U\\u^2=100^2+125^2-2(100)(125)\cos 110^\circ\\u^2=34175.50\\u=184.87$ km[/tex]
Plane Y leaves kano airport at 10.00am and arrives at 11.30am
Time taken =1.5 hour
Therefore:
Average Speed of Y
[tex]=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)[/tex]
b)Flight Direction of Y
Using Law of Sines
[tex]\dfrac{t}{\sin T} =\dfrac{u}{\sin U}\\\dfrac{125}{\sin T} =\dfrac{184.87}{\sin 110}\\123 \times \sin T=125 \times \sin 110\\\sin T=(125 \times \sin 110) \div 184.87\\T=\arcsin [(125 \times \sin 110) \div 184.87]\\T=39^\circ $ (to the nearest degree)[/tex]
The direction of flight Y to the nearest degree is 39 degrees.
