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
Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
Show that none of the following is an identity:
`sin^2 theta + sin theta =2`
If cos A + cos2 A = 1, then sin2 A + sin4 A =
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
Prove that:
tan (55° + x) = cot (35° – x)
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
If 5 tan β = 4, then `(5 sin β - 2 cos β)/(5 sin β + 2 cos β)` = ______.
