From a point hundred metre above the surface of the lake the angular elevation of the peak of Mountain is found to be 30° and the angle of depression of image of the peak is 45° find the height of the peak class 10th
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Answered by
21
This is ur answer hope it will help u
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HarshSharma2004:
i found it helpful
hmmm
will you please tell me how you have taken 100M inside the lake
in the diagram
we are taking angles above 100m to the lake surface
the real image formed has the same height when reflection is seen in mirror
ok
what is m there in 100m +h
the h is the extra height above 100 meter so total height = 100+h
ok thanx
Answered by
4
Sorry could not attach the diagram but it looks something like this..
/ C
/ !
/ !
/ !
/ !
/ !
A/__________! B
! !
! ! [ angle BAC = 30 degree , angle BAE = 45
! ! degree]
! !
____________________
! D
!
!
! E [ join AE diagonally ]
In Triangle BAE
bd/ab=tan 45 = 1
bd=ab
therefore, ab=100m --------------------------(i)
In Triangle ABC
bc/ab=tan30=1/root3
bc=100/root3 {from - (i) }
bc=57.80m
cd=bc+bd
cd=57.80m+100m
cd= 157.80m
/ C
/ !
/ !
/ !
/ !
/ !
A/__________! B
! !
! ! [ angle BAC = 30 degree , angle BAE = 45
! ! degree]
! !
____________________
! D
!
!
! E [ join AE diagonally ]
In Triangle BAE
bd/ab=tan 45 = 1
bd=ab
therefore, ab=100m --------------------------(i)
In Triangle ABC
bc/ab=tan30=1/root3
bc=100/root3 {from - (i) }
bc=57.80m
cd=bc+bd
cd=57.80m+100m
cd= 157.80m
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