Question

An observer flying a aeroplane at an altitude
of 900 mts observe two ships in front of
him which are in the same direction at an
angles of depression of 60² & 300 respectively
find
distance between two ships.​

Answer 1

$\sf{\bold{Correct\:Question:-}}$

An observer flying an aeroplane at an altitude of 900 m and observe two ships in front of him which are in the same direction at an angle of depression 60° and 30° respectively, find distances between two ships.

$\sf{\bold{Solution:-}}$

[ Refer the attachment for figure ]

Let the -

• BC = x
• CD = y (distance between two ships)

Given that AB = 900 m.

In ∆ABC

→ tan60° = AB/BC

→ √3 = 900/x

→ x = 900/√3

→ x = 900/√3 × √3/√3

→ x = 900√3/3

→ x = 300√3 ....(1)

Similarly, In ∆ABD

→ tan30° = AB/BD

→ 1/√3 = AB/(BC + CD)

→ 1/√3 = 900/(x + y)

→ 1/√3 = 900/(300√3 + y)

Cross-multiply them

→ 300√3 + y = 900√3

→ y = 900√3 - 300√3

→ y = √3(900 - 300)

→ y = 600√3

•°• Distance between two ships is 600√3 m

Answer 2

$\huge{\bold{\underline{\underline{\tt Correct \: question:-}}}}$

An observer flying a aeroplane at an altitude of 900 mts observe two ships in front of him which are in the same direction at an angles of depression of 60° & 30° respectively find the distance between two ships.

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let the one ship be at position "C",
• Let the other Ship be at position "D",
• The distance BC = x cm,
• The distance BD = y cm.

(#refer the attachment for the figure),

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

In Δ ABC,

Applying The Trigonometric Functions,

$\large{\boxed{\tt \tan \theta = \dfrac{AB}{DC}}}$

Substituting the values,

$\large{\tt \tan 60 = \dfrac{900}{x}}$

As tan60° = √3,

$\large{\tt \sqrt{3} = \dfrac{900}{x}}$

$\large{\tt x = \dfrac{900}{\sqrt{3}}}$

Rationalizing,

$\large{\tt x = \dfrac{900}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}}$

$\large{\tt x = \dfrac{900 \sqrt{3}}{3}}$

$\large{\tt x = \dfrac{\cancel{900} \sqrt{3}}{\cancel{3}}}$

$\large{\boxed{\tt x = 300 \sqrt{3} \: meters}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

In Δ ABD,

Applying The Trigonometric Functions,

$\large{\boxed{\tt \tan \theta = \dfrac{AB}{DB}}}$

Substituting the values,

$\large{\tt \tan 30 = \dfrac{900}{y}}$

As tan30° = 1/√3,

$\large{\tt \dfrac{1}{\sqrt{3}} = \dfrac{900}{y}}$

$\large{\tt y = 900 \times \sqrt{3}}$

$\large{\boxed{\tt y = 900 \sqrt{3} \: meters}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

The distance between the Two ships is "CD",

To calculate "CD",

$\large{\boxed{\tt CD = BD - BC}}$

Substituting the values,

$\large{\tt CD = y - x}$

$\large{\tt CD = 900 \sqrt{3} - 300 \sqrt{3}}$

$\large{\tt CD = (900 - 300) \times \sqrt{3}}$

$\large{\tt CD = 600 \times \sqrt{3}}$

$\huge{\boxed{\boxed{\tt CD = 600 \sqrt{3} \: meters}}}$

So, The distance between two ships is 600√3 meters .

$\rule{300}{1.5}$