Question

As observed from the top of a light house, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, changes form 30° to 60°. Find the distance travelled by the ship during the period of observation. (Use 3 = 1.73)​

Answer 1

Diagram:

$\setlength{\unitlength}{1cm} \begin{picture}(6,6) \put(2, 2){\line(0, 1){5}} \put(2, 2){\line(1,0){5}} \put(7,2){\line( - 1,1){5}} \put(2,7){\vector(1,0){3}} \put(2,1.7){\vector(1,0){2}} \put(2,1.7){\vector( - 1,0){0.2}}\put(5,1.7){\vector(1,0){2}} \put(5,1.7){\vector( - 1,0){0.2}} \put(2,7){\line(1, - 2){2.5}} \put(1.7, 2){\sf B} \put(4,2.1){\sf C} \put(7,2.1){\sf D} \put(2,7.1){\sf A} \put(3,1.4){\sf y} \put(6,1.4){\sf x} \put(3.7,2.3){ {60}^{\circ} } \put(6,2.3){ {30}^{\circ} }\put(2.3,6.7){ {30}^{\circ} }\put(2.5,6){ {60}^{\circ} } \put(1,4){\sf 100m} \end{picture}$

Given:

• Top of light House is 100m above sea level.
• Angle of depression of a ship Changing from 30° to 60°

Find:

• Distance Travelled by ship during this observation.

Solution:

Let, AB be the height of light house 100m

Let, BC = y and CD = x

$\sf In \: ABC, \dfrac{AB}{BC} = \tan {60}^{\circ}$

$\sf \implies \dfrac{100}{y} = \sqrt{3}$

$\sf \implies y\sqrt{3} = 100$

$\sf \implies y = \dfrac{100}{\sqrt{3}} ....(i)$

_________________________

$\sf In \: ABD, \dfrac{AB}{BD} = \tan {30}^{\circ}$

$\sf \implies \dfrac{100}{y + x} = \dfrac{1}{\sqrt{3}}$

$\sf \implies x + y = 100\sqrt{3}$

$\sf \implies x= 100\sqrt{3} - y.....(ii)$

_________________________

Using eq(i) in eq(ii)

$\sf \dashrightarrow x= 100\sqrt{3} - y$

$\sf \dashrightarrow x= 100\sqrt{3} - \dfrac{100}{ \sqrt{3} }$

$\qquad$ ☯Taking L.C.M☯

$\sf \dashrightarrow x=\dfrac{100 \times 3 - 100}{ \sqrt{3} }$

$\sf \dashrightarrow x=\dfrac{300 - 100}{ \sqrt{3} }$

$\sf \dashrightarrow x=\dfrac{200}{ \sqrt{3} }$

$\qquad$ ☯Rationalising The Denominator☯

$\sf \dashrightarrow x=\dfrac{200}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\sf \dashrightarrow x=\dfrac{200 \sqrt{3} }{3}$

$\qquad$ ☯Using given value of √3 = 1.73☯

$\sf \dashrightarrow x=\dfrac{200 \times 1.73}{3}$

$\sf \dashrightarrow x=\dfrac{346}{3}$

$\sf \dashrightarrow x=115.33m$

Hence, Distance Travelled by the ship is 115.33m