plzz answer the attachment
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Let h be the height of the hill and x m be the distance between the foot of the hill and foot of the tower.
In rt.△ABC,
Cos 60° = x/h
x = h cot 60°........(i)
In rt.△DBC,
cot 30° = x/50
x = 50 cot 30°........(ii)
Equating (i) and (ii)
h cot 60° = 50 cot 30°
h = 50 cot 30°/cot 60°
h = 50 × √3/1√3 = 50 × 3 = 150 m
Therefore the height of the hill is 150 m.
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