Question

5. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of
depression of two ships are 30° and 45°. If one ship is exactly behind the other on the
same side of the lighthouse, find the distance between the two ships,​

Answer 1

Answer:75($\sqrt{3}$-1)

Step-by-step explanation:

let the ships be C and D

and CD = h   (distance between two ships)

In ΔABC

$\frac{75}{BC}=$tan45°

$\frac{75}{BC}=1$

$75 =BC$

In ΔABD

$\frac{75}{75+h} =$tan30°

$\frac{75}{75+h} =\frac{1}{\sqrt{3} }$

$75\sqrt{3} =75+h$

$75\sqrt{3} -75=h$

$75(\sqrt{3}-1)=h$

Answer 2

Given :

It is observed that from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. One ship is exactly behind the other on the same side of the lighthouse.

To Find :

The distance between the two ships.

Solution :

Let DC be the distance between the two ships = x metres.

Let A be the position of the observer at an altitude of 75 m.

then AB = 75 m.

Let C & D be the positions of the two ships whose angles of depression as observed from A are 30° & 45° respectively, then

• ∠ACB = 30°
• ∠ADB = 45°

• ∠ABD = 90°

From right angled ∆ABD,

$\boxed{\bf\tan\ 45^{\circ}=\dfrac{Height}{Base}}$

$\\ \implies\sf\tan\ 45^{\circ}=\dfrac{AB}{BD}$

where,

• tan 45° = 1
• Height = 75 m
• Base = BD m

Substituting the values,

$\\ \implies\sf1=\dfrac{75}{BD}$

By cross multiplying,

$\\ \implies\sf BD=75$

$\\ \therefore\boxed{\bf BD=75\ m.}$

Now,

• DC = x m
• BD = 75 m

So,

• BC = BD + DC = (75 + x) m

From right angled ∆ABC,

$\boxed{\bf\tan\ 30^{\circ}=\dfrac{Height}{Base}}$

$\\ \implies\sf\tan\ 30^{\circ}=\dfrac{AB}{BC}$

where,

• tan 30° = 1/√3
• Height = 75 m
• BC = (75 + x) m

Substituting the values,

$\\ \implies\sf\dfrac{1}{\sqrt{3}}=\dfrac{75}{75+x}$

By cross multiplying,

$\\ \implies\sf(75+x)\times1=75\times\sqrt{3}$

$\\ \implies\sf75+x=75\sqrt{3}$

$\\ \implies\sf x=75\sqrt{3}-75$

Taking √3 as common,

$\\ \implies\sf x=75(\sqrt{3}-1)$

$\\ \implies\sf x=75(1.732-1)$

$\\ \implies\sf x=75(0.732)$

$\\ \implies\sf x=75\times0.732$

$\\ \implies\sf x=54.9$

$\\ \therefore\boxed{\bf x=54.9\ m.}$

The distance between the two ships is 54.9 m.

Explore More :

Trigonometric Ratios :

$\\ \bullet\ \sf\sin\theta=\dfrac{Height}{Hypotenuse}$

$\\ \bullet\ \sf\cos\theta=\dfrac{Base}{Hypotenuse}$

$\\ \bullet\ \sf\tan\theta=\dfrac{Height}{Base}$

$\\ \bullet\ \sf\cot\theta=\dfrac{Base}{Height}$

$\\ \bullet\ \sf\sec\theta=\dfrac{Hypotenuse}{Base}$

$\\ \bullet\ \sf\cosec\theta=\dfrac{Hypotenuse}{Height}$

Trigonometric Table :

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$