An Aeroplane X Whose Average Speed Is 500km/h Leaves Kano Airport At 7:00am And Travels For 2 Hours On A Bearing 050 Degree. It Then Changes Course And Flies On A Bearing 120 Degree To An Airstrip A.
Another Aeroplane Y Leaves Kano Airport At 10:00am And Flies On A Straight Course To The Airstrip A.
Both Planes Arrives At The Airstrip A At 11:30am Calculate
(a) The Average Speed Of Y To Three Significant Figure
(b) The Direction Of Flight Y To The Nearest Degree.

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  1. So what is your question? Do you know what bearing means (000 is North, then090 is East)?

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  2. Draw a diagram, as usual.
    using the law of cosines, the distance z from Kano to Airstrip is
    z^2 = 1000^2 + 1500^2 - 2(1000)(1500) cos110°
    z = 2068

    (a) 2068/1.5
    (b) using the law of sines, the angle between X's starting heading and Y's heading is
    sinθ/1250 = sin110°/2068

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  3. The angle between them is59

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  4. the question asked the average speed, sorry but you've not found the speed I'm not seeing the answer to the average speed there.

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  5. The average speed of Y is 1230km/hr
    Contact me via WhatsApp to get the solutions... 09056199196

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  6. Cant make a drawing to show.... So i hope u get it with this
    X: 7:00am
    Speed of X = 500km/hr for 2hrs
    Diatance of X = 1000km i.e. (500×2)km
    X change course after 2hrs so time will be 9:00am
    X and Y arrive at A at 11:30am
    For X and Y to reach A it will take
    X to reach for 2.5hrs then distance will be (1250km)
    X to A 500km/hr × 2.5hrs = 1250km

    While y will be 1.5hrs since Y left by 10:00am
    Using cosine rule
    |KA|^2 = |KC|^2 + |CA|^2 - 2(|KC||CA|)cos110°
    |KA|= 1848.6km
    Average speed of y = 1848.6/1.5 = 1232.4km
    Approximately to 3.s.f = 1230km

    Direction of flight Y
    Using sine rule
    sin110/ 1848.6 = sinx/1250

    X = 39.5°
    From the starting point the direction of flight will be 50° + 39.5° = 89.5°
    To the nearest degree = 90°

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