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 P (Kano Airport) to point Q in the diagram.
Distance = Speed X Time
Therefore: PQ =50km/hr X 2 hr =100 km
It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.
Distance, QA=50km/hr X 2.5 hr =125 km
Using alternate angles in the diagram:

(a)First, we calculate the distance traveled, PA by plane Y.
Using Cosine rule

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am
Time taken =1.5 hour
Therefore:
Average Speed of Y

(b)Flight Direction of Y
Using Law of Sines
![\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)](https://tex.z-dn.net/?f=%5Cdfrac%7Bp%7D%7B%5Csin%20P%7D%20%3D%5Cdfrac%7Bq%7D%7B%5Csin%20Q%7D%5C%5C%5Cdfrac%7B125%7D%7B%5Csin%20P%7D%20%3D%5Cdfrac%7B184.87%7D%7B%5Csin%20110%7D%5C%5C123%20%5Ctimes%20%5Csin%20P%3D125%20%5Ctimes%20%5Csin%20110%5C%5C%5Csin%20P%3D%28125%20%5Ctimes%20%5Csin%20110%29%20%5Cdiv%20184.87%5C%5CP%3D%5Carcsin%20%5B%28125%20%5Ctimes%20%5Csin%20110%29%20%5Cdiv%20184.87%5D%5C%5CP%3D39%5E%5Ccirc%20%24%20%28to%20the%20nearest%20degree%29)
The direction of flight Y to the nearest degree is 39 degrees.
The second one is the correct equation. The constant to the right should be the diameter squared, while the numbers inside the parentheses should be the opposite.
Answer: Um It’s C?
Step-by-step explanation:
Answer:
If this is FLVS every lesson has I think its called "Function question" underneath is a question. Thoes questions are for the dba. There are questions at the top of every page of every lesson. Look uo the answers for them! or use the notes most of the time they give you answer to them already. I hope that helps of you have any other questions feel free to ask <3