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
Prove the following trigonometric identities.
`(1 + cos theta + sin theta)/(1 + cos theta - sin theta) = (1 + sin theta)/cos theta`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?
(sec A + tan A) (1 − sin A) = ______.
Prove the following identity :
`cosec^4A - cosec^2A = cot^4A + cot^2A`
Choose the correct alternative:
sec2θ – tan2θ =?
Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2
If sin A = `1/2`, then the value of sec A is ______.
