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Question
From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 3 m from the banks, find the width of the river. (Use `sqrt(3)` = 1.73)
Solution
Let the width of the river be x m
i.e., AB = x m
P is the point on the bridge
∴ PQ = 3 m
and ∠RPA = 30° and ∠SPB = 45°
In ΔAPQ, ∠Q = 90° ∠A = 30°
tan A = `(PQ)/(AQ)`
tan 30° = `3/(AQ)`
`1/sqrt(3) = 3/(AQ)`
∴ AQ = `3sqrt(3)` m ...(1)
In ΔPQB, ∠Q = 90° ∠B = 45°
tan B = `(PQ)/(BQ)`
tan 45° = `3/(BQ)`
1 = `3/(BQ)`
BQ = 3 ...(2)
AB = AQ + BQ
= `3sqrt(3) + 3`
= `3(sqrt(3) + 1)` m
Width of River = `3(sqrt(3) + 1)`
= 3 × 2.73
= 8.19 m.
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