bearing of a tower are taken from two points.A and B 250m apart along a straight north-south path, A being due north of B. the bearing of the tower from A and B are 140* and 110* respectively. what is the distance from A and B to the tower.

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  1. I see a triangle ABT, with AB = 250, angle BAT = 40° and angle ABT = 110° making angle T = 30°

    By the Sine Law
    From B to T
    BT/sin40 = 250/sin 30
    BT = 250sin40/sin30 = 321.39 m

    find AT the same way.

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