please solve this question?
This is 4 marks question please solve in detail....
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hello...
In ∆ DCB
tan 60 = DC/ BC
√3 = 50 / BC
BC = 50/√3 m ---------1
Now taking ∆ ABC,
tan 30 = AC/ BC
1 / √3 = h / 50/√3.(by eq.1 )
h = 50 /3 m .
I hope it help you......
in picture u see the figure only solution is wrong ...
I hope it help you.....
In ∆ DCB
tan 60 = DC/ BC
√3 = 50 / BC
BC = 50/√3 m ---------1
Now taking ∆ ABC,
tan 30 = AC/ BC
1 / √3 = h / 50/√3.(by eq.1 )
h = 50 /3 m .
I hope it help you......
in picture u see the figure only solution is wrong ...
I hope it help you.....
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Aasthakatheriya1:
listen
i dont understand what the height of tower is 50m but i understand it as 5 cm.
so i hope u understand
but dont worry it is solve as a sme pattern but put 50 in the place of 5
ok
i understand
thanks
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