Question

From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be30°and45°. Find the distance of the two stones from the foot of the hill. ​

Answer 1

$\sf SOLUTION:$

• 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,

In ∆ABC,tan 45° = AB / AC = P/ B

In ∆ABC,tan 45° = AB / AC = P/ B1 = h / x

• h = x…………………………(1)

In ∆DAB,

$\color{purple}\sf tan 30° = AB / AD = P/ B$

$\tt 1/√3 = h / (x +1)$ [ tan 30° = 1/√3]

$\color{black}\tt √3 h = x + 1$ …………………………(2)

$\color{purple}\tt√3 h = h + 1$ [from eq 1, x = h]

$\color{purple}\rm √3 h - h = 1$

$\color{purple}\sf h(√3 -1) = 1$

$\color{purple}\sf h = 1/ (√3 -1)$

$\color{purple}\sf h = 1/ (√3 -1)$

$\color{navyblue}\sf = 1× (√3 + 1) / (√3 -1) (√3 +1)$

[ By rationalising ]-:

$\color{red}\sf h = (√3 + 1) / √3² - 1² = (√3 + 1) / 3 - 1$

$\color{purple}\sf h = (√3 + 1) / 2$

$\color{purple}\sf h = (1.73 +1)/2$ [√3 = 1.73]

$\color{purple}\sf h = 2.73/2 = 1.365$

$\color{purple}\sf h = 1.365 km$

• Hence,the height of the hill is 1.365 km.

Answer 2

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,

In ∆ABC,tan 45° = AB / AC = P/ B

In ∆ABC,tan 45° = AB / AC = P/ B1 = h / x

h = x…………………………(1)

In ∆DAB,

tan30° = AB/AD

1/√3 = h/(x+1)

√3h = x + 1....(2)

√3=h+1

√3h-h = 1

h(√3-1) = 1

h = 1/√3+1

=1×(√3+1)/(√3−1)(√3+1)

By rationalising:

h=(√3+1)/√3²−1²=(√3+1)/

h=(√3+1)/2

h = (1.73 +1)/2

h=2.73/2=1.365

h = 1.365 km

Hence,the height of the hill is 1.365 km.

Hope it help yoh mate !!