Q20 of 47 Page 1

A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h.

The figure is given as:



Given: AB = Height of cliff = 150 m


Let BC =x m


Let CD = y m


Then on ΔABC,


tan 60° = AB/BC


⇒ √3 = 150/x


⇒ x = 150√3 ………... (1)


In ΔABD,


tan 45° = AB/BD


⇒ 1 = 150/(x+y)


⇒ x+y = 150 ………………(2)


From equation (1) and (2), we get:


y = 150 –x = 150 – (150/√3)


= 150 – (150√3/3) (multiplying and dividing the later term by √3)


= (150 – 50√3)


Now, time taken by the boat to move from point C to point D = 2 min = 2/60 hr = 1/30 hr


Distance = Speed × Time


∴ Speed = Distance/Time


= (150 – 50√3)/(1/30)


= 30 × (150 – 50√3)


= (4500 – 1500√3) m/hr


= 1901.92 m/hr


= 1.90192 km/hr


More from this chapter

All 47 →
18

The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is (7/2, y); find the value of y.

 19

Two different dice are thrown together. Find the probability that the numbers obtained

(i) have a sum less than 7


(ii) have a product less than 16


(iii) is a doublet of odd numbers.

 21

Construct an isosceles triangle with base 8 cm and altitude 4 cm. Construct another triangle whose sides are 2/3 times the corresponding sides of the isosceles triangle.

 22

Prove that the lengths of tangents drawn from an external point to a circle are equal.