Discussion
A tower stands vertically on the ground. From a point on the ground, which is 17 m away from foot of the tower, the angle of elevation of the top of the tower is found to be 30°. What is the height of the tower ?
Answer: A
The height of the tower is \(17/\sqrt{3}\) m.
The tower, the ground and the line of sight form a right triangle with the 30° angle at the observer, the 17 m distance as the adjacent side and the height h as the opposite side.
\[ \begin{aligned} \tan 30^{\circ} &= \frac{h}{17} \\ h &= \frac{17}{\sqrt{3}} \approx 9.8\ \text{m} \end{aligned} \]
\(17\sqrt{3}\) would be the height for a 60° angle of elevation, and 17 m would need a 45° angle.
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