Advertisements
Advertisements
प्रश्न
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/min.
Advertisements
उत्तर
Let AO be the cliff of height 150 m.
Let the speed of the boat be x meters per minute.
And BC is the distance which man travelled.
So, BC = 2x ....[ ∵Distance = Speed x Time ]
tan(60°) = `"AO"/"OB"`
`sqrt3` = `150/"OB"`
⇒ OB = `(150sqrt3)/3` = `50sqrt3`
tan(45°) = `"AO"/"OC"`
⇒1 = `150/"OC"`
⇒ OC = 150
Now OC = OB + BC
⇒ 150 = `50sqrt3` + 2x
⇒ x = `(150 − 50sqrt3)/2`
⇒ x = 75 − `25sqrt3`
Using `sqrt3 = 1.73`
x = 75 − 25 x 1.732 ≈ 32 m/min
Hence, the speed of the boat is 32 metres per minute.
संबंधित प्रश्न
As observed from the top of an 80 m tall lighthouse, the angles of depression of two ships on the same side of the lighthouse of the horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest meter.
Prove the following trigonometric identities.
`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
Prove the following identities.
`(sin^3"A" + cos^3"A")/(sin"A" + cos"A") + (sin^3"A" - cos^3"A")/(sin"A" - cos"A")` = 2
