a bird sitting on the top of a tree, which is 80m high. the angle of elevation of the bird, from the point on the ground is 45. the bird flies away from the point of observation horizantally and remains at a constant height. after 2 seconds, the angle of elevation of the bird from the point of observation becomes 30. find the sped of the flying bird (use √3=1.7
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In triangle OAC tan 45=AC/OA=1
AC=OA=80
IN TRIANGLE ODB
tan 30=BD/OB=1/√ 3
OB=BD√ 3-AC√ 3(since ac=bd as bird remains constant height)
AB=OB-OA=AC√ 3-AC=AC(√ 3-1)=80(√ 3-1)
NOW FOR 2 SEC
80(√ 3-1)/2=40(√ 3-1)
=29.2M/SEC
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AC=OA=80
IN TRIANGLE ODB
tan 30=BD/OB=1/√ 3
OB=BD√ 3-AC√ 3(since ac=bd as bird remains constant height)
AB=OB-OA=AC√ 3-AC=AC(√ 3-1)=80(√ 3-1)
NOW FOR 2 SEC
80(√ 3-1)/2=40(√ 3-1)
=29.2M/SEC
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mark me as Brain list
plzz plzz
plzzs
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