From the top of a 7m 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° .
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Answer
- Height of the tower is 7(√3+1) m
Explanation
Given
- From the top of a 7 m High building the angle of elevation to the top is 60°
- Angel of depression to the foot is 45°
To Find
- Height of the tower
Solution
- Refer the diagram for the Diagram? Here we shall first use the tan function to find the horizontal length first, then use the same on ∆AED to find the height of DE. Then the total height will be DE + EC
✭ In ∆ABC
→ tan θ = Opposite/Adjacent
→ tan 45° = AB/BC
[tan 45°] = 1
→ 1 = 7/y
→ 1 × y = 7
→ y = 7 m
✭ In ∆AED
→ tan θ = Opposite/Adjacent
→ tan 60° = DE/AE
[tan 60° = √3]
→ √3 = x/y
→ √3 = x/7
→ √3 × 7 = x
→ x = 7√3
✭ Height of the tower
→ Height (Tower) = DE+EC
→ Height (Tower) = 7√3 + 7
→ Height (Tower) = 7(√3+1) m
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Answered by
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Answer:
- From the top of a 7m high building
- Angle of elevation of the top of a cable tower is 60°
- Height of tower
From the attachment,
Let AB be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is 60° and the angle of depression of its foot
From ΔABC,
We know that ,Tan45° = 1
We know that, Tan60° =√3
Height of tower :-
EC = DE + CD
= 7√3 + 7
= 7(√3 + 1 )
Explanation:
Therefore,Height of tower = 7 (√3 + 1) m
Attachments:
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