Discussion
A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60°
Answer: A
Let the tower be \( h \) m and the observer \( d \) m away. The flagpole's bottom is the tower's top.
\[ \tan 30^{\circ} = \frac{h}{d} \quad\Longrightarrow\quad d = h\sqrt{3} \]
\[ \tan 60^{\circ} = \frac{h+18}{d} \quad\Longrightarrow\quad h + 18 = \sqrt{3} \cdot h\sqrt{3} = 3h \]
\[ 2h = 18 \quad\Longrightarrow\quad h = 9 \text{ m} \]
Since \( \tan 60^{\circ} \div \tan 30^{\circ} = 3 \), the tower-plus-flagpole is always three times the tower alone - so the flagpole equals twice the tower.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development