Question

From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 45° and 30° respectively. Find the height of the hill. [Take `sqrt(3) = 1.732`.]

Answer 1

Let AB be the hill and let C and D be the two consecutive kilometre stones. Then,

∠ACB = 45°, ∠ADB = 30° and CD = 1 km = 1000 m.

Let AB =h m and AC = x m.

From right ΔBAC, we have

`(AC)/(AB) = cot 45^circ = 1`

⇒ `x/h = 1`

⇒ x = h   ...(i)

From right ΔBAD, we have

`(AB)/(AD) = tan 30^circ = 1/sqrt(3)`

⇒ `h/(x + 1000) = 1/sqrt(3)`

⇒ `h/(h + 1000) = 1/sqrt(3)`   ...[From (i)]

⇒ `(sqrt(3) - 1)h = 1000`

∴ `h = 1000/((sqrt(3) - 1))`

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

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

= (500 × 2.732) m

= 1366 m