A smuggler leaves a private airfield at 06:00 flying at a heading of N40E at 200km/h. The plane is detected by radar at the police airport, which is located 150km northwest of the private airfield. At 06:30 the police airplane leaves its airport with the intention of intercepting the smuggler at 08:30. Determine the course (direction) and speed of the police airplane to the nearest unit.

Direction of the police plane = N57.1E

Speed of the police airplane = 255 km/h

Step-by-step explanation:

The diagram of the situation described is presented in the attached image to this question.

Let the distance the police airplane has to travel to intercept the smuggler at 08:30 be x km

The police airplane moves at 06:30 and plans to intercept the smuggler at 08:30; thereby travelling for 2 hours.

By 08:30, the smuggler would have travelled for 2 hours 30 mins, that is, 2.5 hours, travelling at 200 km/h, that is a total distance of 500 km covered.

So, the paths form a triangle.

Using cosine rule, we can obtain the distance, x, that the police airplane has to travel to intercept the smuggler at 08:30.

x² = 150² + 500² - (2×150×500×cos 85°)

x² = 259,426.63858785

x = 509.34 km

We can obtain the direction, Φ, by finding the angle θ using some rule.

[(Sin 85°)/509.34] = [(Sin θ)/500]

Sin θ = (500 × sin 85°)/509.34 = 0.9779 = 77.94°

From the attached image,

Φ + θ = 90° + 45° = 135°

Φ = 135° - θ = 135° - 77.94° = 57.06° = 57.1°

Therefore,

Speed of police airplane = (distance)/(time) = (509.34/2) = 254.67 km/h = 255 km/h

Direction of the police plane = N57.1E

Hope this Helps!!!

3

step-by-step explanation:

answer: the proof is mentioned below.

step-by-step explanation:

since, given:   lines bc = ef

prove:   lines ac > ef

          statements                                           reason

1. bc=ef                                           1. given

2. ac= ab+bc                                 2. because b is the point between the  

                                                            line segment ac

3.   ac > bc                                       3. because if a line segment is equal to

                                                          the sum of two line segments then it is

                                                          always greater then each of these

                                                          individual line segments

4. ac > ef                                         4. by equality property.