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प्रश्न
The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.
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उत्तर

Let PQ be the light house.
`=>` PQ = 60
`tan 60^@ = (PQ)/(AQ)`
`=> sqrt(3) = 60/(AQ)`
`=> AQ = 60/sqrt(3)`
`=> AQ = (20 xx 3)/sqrt(3)`
`=> AQ = (20 xx sqrt(3) xx sqrt(3))/sqrt(3)`
`=> AQ = 20sqrt(3) m`
In ΔPQB
`tan 45^@ = (PQ)/(QB)`
`=> 1 = 60/(QB)`
`=>` QB = 60 m
Now,
AB = AQ + QB
= `20sqrt(3) + 60`
= 20 × 1.732 + 60
= 94.64
= 95 m
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(cos^2 theta)/sin theta - cosec theta + sin theta = 0`
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`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
