Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answers (1)

best_answer


Let the height of the cable tower be (AB = h+7)m
Given,
Height of the building is 7 m and angle of elevation of the top of the tower \angle ACE = 60^o , angle of depression of its foot \angle BCE = 45^o.

According to question,

In triangle \Delta DBC,
\\\tan 45^o = \frac{CD}{BD} = \frac{7}{BD} = 1\\ BD =7\ m
since DB = CE = 7 m

In triangle \Delta ACE,

\\\tan 60^o = \frac{h}{CE}=\frac{h}{7}=\sqrt{3}\\ \therefore h = 7\sqrt{3}\ m

Thus, the total height of the tower equal to h+7 =7(1+\sqrt{3}\) m

Posted by

manish

View full answer

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 10, 12 syllabus 2024-25 out; board restores two-level mathematics rule
anam-khan

CBSE Class 10 board exam 2024 passing criteria, eligibility for compartment; all you need to know

Latest Question