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Question
At a particular time, when the sun’s altitude is 30°, the length of the shadow of a vertical tower is 45 m. Calculate:
- the height of the tower.
- the length of the shadow of the same tower, when the sun’s altitude is:
- 45°
- 60°
Solution
Shadow of the tower = 45 m and angle of elevation = 30°
i. Let AB be the tower and BC is its shadow.
∴ CB = 45 m.
Now in right ΔABC,
`tan theta = (AB)/(BC)`
`\implies tan 30^circ = (AB)/45`
`\implies 1/sqrt(3) = (AB)/45`
`\implies AB = 45/sqrt(3)m`.
∴ `AB = (45 xx sqrt(3))/(sqrt(3) xx sqrt(3))`
= `(45sqrt(3))/3`
= `15sqrt(3) m`.
= 15 (1.732)
= 25.980
= 25.98 m
ii. In second case,
a. Angle of elevation = 45°
And height of tower = 25.98 m
or `15sqrt(3) m`
`tan 45^circ = (AB)/(DB)`
`\implies (AB)/(DB) = 1`
∴ DB = AB = 25.98 m.
b. Angle of elevation = 60°
And height of tower = 25.98 m
or `15sqrt(3) m`.
Let shadow of the tower DB = x m
∴ `tan 60^circ = (AB)/(DB)`
`\implies sqrt(3) = (15sqrt(3))/(DB)`
`\implies DB = (15sqrt(3))/sqrt(3) = 15 m`.
Hence length of shadow = 15 m.
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