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प्रश्न
A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°.
Calculate :
- the width of the river;
- the height of the tree.
उत्तर १
Let AB be the tree and AC be the width of the river.
Let D be a point such that CD = 50 m.
Given that ∠BCA = 60° and ∠BDA = 30°
In ΔBAD,
`(BA)/(AD) = tan 30^circ`
`=> BA = (AD)/sqrt(3)` ...(i)
In ΔBAC,
`(BA)/(AC) = tan 60^circ`
`=> BA = ACsqrt(3)` ...(ii)
From (i) and (ii), we get
`(AD)/(sqrt(3)) = ACsqrt(3)`
`=>` (50 + AC) = 3AC
∴ AC = 25 m
Thus, width of the river is 25 m.
From (ii),
BA = 25 × 1.732 = 43.3 m
Hence, height of the tree is 43.3 m.
उत्तर २
Let H be the height of the tree
Let height of the tree be H meter.
In right-angled ΔACD,
`tan 60^circ = H/(CD)`
`sqrt(3) = H/(CD)`
∴ `CD = H/sqrt(3)` ...(i)
In right-angled ΔABD,
`tan 30^circ = H/(BD)`
∴ `1/sqrt(3) = H/(BD)`
`BD = sqrt(3)H` ...(ii)
BD – CD = 50
`(sqrt(3)H)/1 - H/sqrt(3) = 50` ...(Using (i) and (ii))
∴ `(3H - H )/sqrt(3) = 50`
∴ `2H = 50sqrt(3)`
or `H = (50sqrt(3))/2 = 25sqrt(3)`
H = 43.3 m
i. The width of the river `CD = (25sqrt(3))/sqrt(3) = 25 m`
ii. The height of the tree H = 43.3 m.
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