an aircraft is flying along a horizontal path PQ directly towards an observer on the ground at O and maintaining an attitude of 300m at when the aircraft is at P the angle of depression is 30o when at Q the angle of depression is 60o find the distance PQ
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In this question we use trigonometric ratios to solve.
First you put this movement in a diagram showing the altitude where the aircraft is and the respective angles of depression.
In this question since we have two angles of depressions at different points, then it implies that we will have two right angles with the same height and sharing a base which is PQ.
You let the length of base of the triangle with angle of depression as 30° be x1 and the other one x2.
In this case we have the opposite sides and the adjacent sides, so we use Tan ratio.
Tan 30°=x1/300m
x1=300Tan 30°=173.21m
x2=300Tan 60°=519.62m
PQ=x1 + x2
173.21 + 519.62=692.83m
First you put this movement in a diagram showing the altitude where the aircraft is and the respective angles of depression.
In this question since we have two angles of depressions at different points, then it implies that we will have two right angles with the same height and sharing a base which is PQ.
You let the length of base of the triangle with angle of depression as 30° be x1 and the other one x2.
In this case we have the opposite sides and the adjacent sides, so we use Tan ratio.
Tan 30°=x1/300m
x1=300Tan 30°=173.21m
x2=300Tan 60°=519.62m
PQ=x1 + x2
173.21 + 519.62=692.83m
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