CBSE Class 10 Answered
heights and distances
Asked by | 29 Sep, 2008, 06:04: PM
Expert Answer
(Pl see attached figure)
Let the poin5t of observation be A , initial position of the plane be P and new position be Q . BP = CQ = 3000m
Angle PAB = 45 deg and Angle QAC = 30 deg
Consider triangle PBA . Since one angle is 90 deg and the other is 45deg , the third angle is also 45 deg , so AB = PB = 3000m
Now let BC = x . In triangle AQC 3000/ ( 3000+x) = tan 30 = 1/ 3
therefore , 30003 = 3000+x
3000(3 -1) = x
3000 x 0.732 = x
x= 2196m
speed = distance (in km)/ time (in hrs)
=2196 x 3600 / 15 x 1000 = 527 . 04 km / hr
Answered by | 30 Sep, 2008, 11:57: AM
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