Question

plz answer this in the attachment

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Answer 1

Answer:

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.

Answer 2

Answer:

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.