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
Solution:
Let the height of the tower be CE and the height of the building be AB. The angle of elevation from the top E of the tower to the top A of the building is 60° and the angle of depression from the bottom C of the tower to the top A of the building is 45°.
Draw AD || BC.
Then, ∠DAC = ∠ACB = 45° (alternate interior angles)
In ΔABC,
tan 45° = AB / BC
1 = 7/BC
BC = 7
ABCD is a rectangle,
Therefore, BC = AD = 7 and AB = CD = 7
In ΔADE,
tan 60° = ED/AD
√3 = ED/7
ED = 7√3
Height of tower = CE = ED + CD
= 7√3 + 7
= 7 (√3 + 1)
Height of the tower = 7 (√3 + 1) m.
☛ Check: NCERT Solutions Class 10 Maths Chapter 9
Video Solution:
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
Maths NCERT Solutions Class 10 Chapter 9 Exercise 9.1 Question 12
Summary:
If 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°, then the height of the tower is 7(1 + √3) m.
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