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Question
The angle of elevation of the top of a 100 m high tree from two points A and B on the opposite side of the tree are 52° and 45° respectively. Find the distance AB, to the nearest metre.
Solution
In ΔADC,
tan 52° = `"DC"/"AC" = 100/"AC"`
`\implies` 1.2799 = `100/"AC"` ...(From table)
`\implies` AC = `100/1.2799`
`\implies` AC = 78.13 m
In ΔBCD,
tan 45° = `"CD"/"BC"`
`\implies` 1 = `100/"BC"`
BC = 100 m
∴ AB = AC + BC
= 78.13 + 100
= 178.13 m
Hence, the distance AB is 178 m ...(approx)
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