Question

From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be 30° and 45° respectively. Find the distances of the two stones from the foot of the hill.

Answer 1

Let AB be the hill of height 'h' km and C and D be two consecutive stones such that CD = 1 km, ∠ACB = 30° and ∠ADB = 45°

In ΔABD,

`(AB)/(BD) = tan 45^circ = 1`

`=>` BD = h

In ΔABC,

`(AB)/(BC) = tan 30^circ`

`=> h/(BC) = 1/sqrt(3)`

`=>  h/(h + 1) = 1/sqrt(3)`

`=> h = 1/(sqrt(3) - 1)`

= `(sqrt(3) + 1)/2`

= `2.732/2`

= 1.366 km

∴ BD = 1.366 km

BC = BD + DC

= 1.366 + 1

= 2.366 km

Hence, the two stone are at a distance of 1.366 km and 2.366 km from the foot of the hill. 

Answer 2

Let AB be hill of which B is foot of hill and D and C are two consecutive Km stones.

∴ DC = 1 km = 1000 m

In right-angled ΔABC,

`tan 45^circ = (AB)/(BC)`

`1 = h/x`

x = h     ...(i)

In right-angled ΔABD,

`tan 30^circ = (AB)/(BD)`

`1/sqrt(3) = h/(x + 1000)` 

`x + 1000 = hsqrt(3)`    ...(ii)

But from equation (i), x = h,

∴ `x + 1000 = xsqrt(3)`

`x(sqrt(3) - 1) = 1000`

`x = 1000/(sqrt(3) - 1) xx (sqrt(3) + 1)/(sqrt(3) + 1)`

`x = (1000(sqrt(3) + 1))/(2)`

`x = 500(sqrt(3) + 1)`

x = 500 × 2.732

x = 1366 metre

x = 1.366 km

∴ 1^st km stone is 1.366 km and 2^nd km stone is 2.366 km from foot of hill.