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Question
A tower is 60 m heigh. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30°, then x is equal to
Options
41.92 m
43.92 m
43 m
45.6 m
Solution
43.92 m
Explanation;
Hint:
In the right ∆ABC, tan 30° = `"AB"/"BC" = 60/(x + y)`
`1/sqrt(3) = 60/(x + y)`
⇒ x + y = `60sqrt(3)`
y = `60sqrt(3) - x` ...(1)
In the right ∆ABD, tan 45° = `"AB"/"BD"`
1 = `60/y`
⇒ y = 60 ...(2)
From (1) and (2) we get
60 = `60sqrt(3) - x`
x = `60 sqrt(3) - 60`
= `60(sqrt(3) - 1)`
= 60(1.732 – 1)
= 60 × 0.732
x = 43.92 m
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