3 Views

Question : The angle of depression of two ships from the top of a lighthouse is $60^{\circ}$ and $45^{\circ}$ towards the east. If the ships are 300 metres apart, the height of the lighthouse (in metres) is:

Option 1: $200\left ( 3+\sqrt{3} \right )$

Option 2: $250\left ( 3+{\sqrt3} \right )$

Option 3: $150\left (3 +{\sqrt3} \right)$

Option 4: $160 \left (3+{\sqrt3} \right)$


Team Careers360 14th Jan, 2024
Answer (1)
Team Careers360 20th Jan, 2024

Correct Answer: $150\left (3 +{\sqrt3} \right)$


Solution :
In $\Delta ABC$
$\tan 60^{\circ} = \frac{AB}{BC}$
$⇒\sqrt{3} = \frac{AB}{BC}$
$⇒AB = \sqrt{3} BC \quad \text{-----------------------(1)}$
In $\Delta ABD$
$\tan 45^{\circ} = \frac{AB}{BD}$
$\Rightarrow 1 = \frac{AB}{BD}$
$⇒AB = BD \quad \text{---------------------------------(2)}$
From equation $(1)$ and equation $(2)$
$\Rightarrow BD = \sqrt{3} BC\quad \text{--------------------------(3)}$
From the figure
$BD = BC + CD$
From equation $(3)$
$\Rightarrow \sqrt{3} BC = BC + 300$
$\Rightarrow BC(\sqrt{3} - 1) = 300$
$\Rightarrow BC = \frac{300}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$
$\Rightarrow BC = \frac{300}{2} \times (\sqrt{3} + 1)$
$\Rightarrow BC = 150 (\sqrt{3} + 1)$
From equation (1)
$AB = 150 (\sqrt{3} + 1) \times \sqrt{3}$
$\therefore AB = 150 (3 + \sqrt{3})$ metres.
Hence, the correct answer is $150 (3 + \sqrt{3})$ metres.

SSC CGL Complete Guide

Candidates can download this ebook to know all about SSC CGL.

Download EBook

Know More About

Related Questions

TOEFL ® Registrations 2024
Apply
Accepted by more than 11,000 universities in over 150 countries worldwide
Manipal Online M.Com Admissions
Apply
Apply for Online M.Com from Manipal University
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books