Question

11.As observed from the top of a light house 100m above the sea level, the angle of depression of a ship sailing directly towards it changes from 30deg to 45deg. Determine the distance travelled by the ship during the period of observation​

Answer 1

Answer:

$\bigstar{\bold{Distance\:travelled=100(\sqrt{3}-1)\:m}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Height of the lighthouse = 100 m
• Initial angle of depression = 30°
• Final angle of depression = 45°

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Distance travelled by the ship during this time period

$\Large{\underline{\underline{\bf{Solution:}}}}$

⇝ Let the height of lighthouse = AB = 100 m

⇝ We have to find the distance travelled by ship = DC

⇝ First consider Δ ABC

    tan 45 = AB/BC

    tan 45 = 100/BC

    1 = 100/BC

    BC = 100 m

⇝ Now consider Δ ABD

    tan 30 = AB/DB

⇝ But we know that,

    DB = DC + CB

⇝ Hence,

    tan 30 = AB/DC + CB

    tan 30 = AB/(DC + 100)

    1/√3 = 100/DC + 100

⇝ Cross multiplying,

    DC + 100 = 100 √3

    DC = 100√3 - 100

    DC = 100 (√3 - 1)

⇝ Hence the distance travelled is 100 (√3 - 1) m

    $\boxed{\bold{Distance\:travelled=100(\sqrt{3}-1)\:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

⇝ Sin A = opposite/hypotenuse

⇝ Cos A = adjacent/hypotenuse

⇝ tan A = opposite/adjacent