Question

From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 30° and 45°. Find the height of the hill.

Answer 1

SOLUTION:
Let AB = h km be the height of the hill & C & D be two stones.
Let DC = 1 km & AC = x km.

∠ACB = ∠EBC = 45°  [ alternate angles]
∠ADB = ∠EBD = 30°  [ alternate angles]

In ∆ABC,
tan 45° = AB / AC = P/ B
1 = h / x              [ tan 45° = 1]
h = x…………………………(1)

In ∆DAB,
tan 30° = AB / AD = P/ B
1/√3 = h / (x +1)             [ tan 30° = 1/√3]
√3 h = x + 1 …………………………(2)
√3 h = h + 1             [from eq 1, x = h]
√3 h - h  = 1  
h(√3 -1) = 1
h = 1/ (√3 -1)
h = 1/ (√3 -1)
= 1×  (√3 + 1) /  (√3 -1) (√3 +1)

[ By rationalising ]

h = (√3 + 1) / √3² - 1² = (√3 + 1) / 3 - 1
h = (√3 + 1) / 2
h = (1.73 +1)/2          [√3 = 1.73]
h = 2.73/2 = 1.365
h = 1.365 km

Hence,the height of the hill is 1.365 km.

HOPE THIS WILL HELP YOU....

Answer 2

let height of hill h km

first stone at point C at a distance xkm from base of hill
second stone at Point D which 1km away from 1st stone said in Question

in right angled ∆ ABC
h/x = tan45° = 1
h = x

in ∆ right angled ∆ ABD

h/x+1 = tan30° = 1/√3

√3h = x+1
√3h = h+1 ( because x=h)
√3h-h =1
h(√3-1)= 1
√3 = 1.732
h = 1/0.732 = 1.366 km Answer