A person standing on the bank of a river observes that the angle of elevation of the top of the tree standing on the opposite bank is 60 degree .When he moves 30m away from the bank, he finds the angle of elevation to be 30degree. Find the height of the tree and width of the river.
2.The angle of elevation of the top of a tower from a point P and Q at distance of 4m and 9m respectively from the base of the tower and in the same straight line with it are 60degree and 30degree.Prove that the height of the tower is 6m.
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Ans1. In∆ABC, tan 60°=P/B
√3=AB/x
√3x=AB. -(I)
In∆ABD, tan 30°= P/B
1/√3 = AB/ 30+x
By cross multiplication-
30+x=√3h
30+x=√3×√3x (from equation (I)
30+x= 3x
30=3x-x
30=2x
x=15
Now, put x=15 in equation (I)
√3x=h
10√3=h
Therefore, height of the tree is 10√3m and width of the river =30+x
= 30+15
=45m
Hope you will get your answer.
√3=AB/x
√3x=AB. -(I)
In∆ABD, tan 30°= P/B
1/√3 = AB/ 30+x
By cross multiplication-
30+x=√3h
30+x=√3×√3x (from equation (I)
30+x= 3x
30=3x-x
30=2x
x=15
Now, put x=15 in equation (I)
√3x=h
10√3=h
Therefore, height of the tree is 10√3m and width of the river =30+x
= 30+15
=45m
Hope you will get your answer.
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