Question

From the top of a lighthouse, 80 metres high, two ships on the same side of the lighthouse are observed.

The angles of depression of the ships as seen from the lighthouse are found to be of 45° and 30°. Find the

distance between the two ships.(Assume that the two ships and the bottom of the lighthouse are in a line.​

Answer 1

Answer:

• Distance between two ships = 80 (√3 - 1) metres

Explanation:

Given:

• Height of Lighthouse = 80 m
• There are two ships on same side of lighthouse, the angles of depression of the ships are found to be 45° and 30°

To find:

• Distance between the two ships

Refer to the figure attached:

• AB is the lighthouse of height 80 m with B as base.
• D and C are two boats  with angle of elevation of lighthouse being 30° and 45° respectively.
• CD is the distance between Two boats so, CD =?

Solution:

In Δ ACB

→ tan 45° = AB / BC

( using value of tan 45° = 1 )

→ 1 = 80 / BC

→ BC = 80  m

Now,

In Δ ADB

→ tan 30° = AB / BD

( using tan 30° = 1 / √3 )

→ 1 / √3 = 80 / BD

→ BD = 80√3  m

Further,

→ CD = BD - BC

→ CD = 80√3 - 80

→ CD = 80 (√3 - 1)  m

Therefore,

• Distance between two ships is 80 (√3 - 1) metres.

Answer 2

Answer:

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(8,1){\line(1,0){4}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\qbezier(12,1)(11,1.4)(8,2.9)\put(7.3,2){\sf{\large{80 m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\qbezier(11,1)(10.8,1.2)(11.1,1.4)\put(9.4,1.2){\sf\large{45^{\circ} }}\put(10.5,1.2){\sf\large{30^{\circ} }}\put(11.9,.7){\sf\large D}\put(7.9,3){\sf\large A}\put(10.4,.7){\sf\large B}\put(7.9,.7){\sf\large C}\end{picture}$

• In Triangle ABC :

$:\implies\sf tan(\theta)=\dfrac{Perpendicular}{Base}\\\\\\:\implies\sf tan(45^{\circ})=\dfrac{80}{BC}\\\\\\:\implies\sf 1=\dfrac{80}{BC}\\\\\\:\implies\sf BC=\dfrac{80}{1}\\\\\\:\implies\sf BC=80\qquad -eq.(I)$

$\rule{180}{1.5}$

• In Triangle ACD :

$:\implies\sf tan(\theta)=\dfrac{Perpendicular}{Base}\\\\\\:\implies\sf tan(30^{\circ})=\dfrac{80}{CD}\\\\\\:\implies\sf \dfrac{1}{ \sqrt{3} }=\dfrac{80}{BC+BD}\\\\\\:\implies\sf BC+BD=\dfrac{80}{\frac{1}{ \sqrt{3} }}\\\\\\:\implies\sf 80+BD=80 \sqrt{3}\\\\\\:\implies\sf BD = 80 \sqrt{3} - 80\\\\\\:\implies\underline{\boxed{\sf BD = 80( \sqrt{3} - 1) \:metres}}$