Advertisements
Advertisements
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.
Advertisements
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
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities:
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities.
`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`
Prove the following identities:
`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
Show that : tan 10° tan 15° tan 75° tan 80° = 1
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
`{1/((sec^2 theta- cos^2 theta))+ 1/((cosec^2 theta - sin^2 theta))} ( sin^2 theta cos^2 theta) = (1- sin^2 theta cos ^2 theta)/(2+ sin^2 theta cos^2 theta)`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
