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Question on Triangles~
Asked by anppn77 | 10 Mar, 2009, 10:11: PM

BD is the string, Dis the fly which is 3.6m away from point A. BC is the vertical dist of the tip of the rod to the water.

In right triangle BCD

(BC)2 + (CD)2 =(BD)2

(1.8)2 + (2.4)2  =(BD)2

9.00 =(BD)2

Therefore the  length  of the string  BD is 3m which is nothing but 300cm.

Now if the string is pulled @ of 5 cm /sec then ,it is pulled 60 cm in 12 sec.which means length BD (string)now becomes 240 cm(300-60) i.e 2.4m.now we must find out dist  AD, of which dist AC remains same.therefore to find CD now

In triangle BCD

(BC)2 + (CD)2 =(BD)2

(1.8)2 + (CD)2  =(2.4)2

(2.4)2  - (1.8)2 =(CD)2

5.76   -3.24 = (CD)2

2.52 = (CD)2

CD =1.587 or 1.6 m

therefore after 12 sec the fly is 2.8 m (1.6 +1.2) m away from Nazima

Answered by | 20 Jul, 2009, 05:46: PM
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