Advertisements
Advertisements
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.
Advertisements
उत्तर
Let AB be the lighthouse and C and D be the two ships.
The angles of depression of the 2 ships are 30° and 40°
So, ∠ADB = 30°
∠ACB = 40°
Let the distance between the ships be CD = x m.
Also, Let BD = y m.
In ΔABC
`tan 40^@ = 80/(y - x)`
`=> y - x = 80/0.8390 = 95.352` ....(1)
Also from ΔABD
`tan 30^@ = 80/y`
`=> y = 80sqrt3 = 80 xx 1.732m` = 138.56 m
From (1) we get
`138.56 - x = 95.352`
`=> x = 138.56 - 95.352 cm = 43.208 m ~~ 43m`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = sec A + tan A`
Prove the following identities:
`cot^2A((secA - 1)/(1 + sinA)) + sec^2A((sinA - 1)/(1 + secA)) = 0`
Define an identity.
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
Prove that `(cot A)/(1 - cot A) + (tan A)/(1 - tan A) = -1`.
Prove that 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0.
If tan θ – sin2θ = cos2θ, then show that `sin^2θ = 1/2`.
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
