WILL AWARD BRAINLIEST! A control tower observes the flight of an aircraft.
At 09:23 the aircraft is 580 km away on a bearing of 043º.
At 09:25 the aircraft is 360 km away on a bearing of 016º.
What is the speed and the course of the aircraft?
(Use a scale of 1 cm to 50 km.)

27465 km/h

Step-by-step explanation:

From the figure, let us use cosine formula to calculate the resultant displacement.

B^2 = C^2 + A^2 - 2(A)(C) cosØ

Where C = 580km

A = 360 km

Ø = 153 degree

Substitute all the parameters into the formula

B^2 = 580^2 + 360^2 - 2(360)(580)cos153

B^2 = 466000 - ( - 372084.32 )

B^2 = 466000 + 372084.32

B^2 = 838084.32

Square root both sides

B = 915.5 km

You are told to use a scale of 1 cm to 50 km.

Therefore, B = 915.5/50 = 18.3 cm

The time given are: 09:23 and 09:25.

The time difference = 25 - 23 = 2 minute.

Convert minutes to hours

2 minute = 2/60 = 1/30 hours

Speed = distance/time

Speed = 915.5 ÷ 1/30

Speed = 915.5 × 30

Speed = 27465 km/h

Or

Convert 2 minute to second

2 minute = 2 × 60 = 120 seconds

Speed = distance/time

Speed = 18.3/120

Speed = 0. 1525 cm/s

i dont know sorry

step-by-step explanation:

98%

step-by-step explanation: