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Is 98 correct?
You did like this no?? Let the angle be A Let the whole length be x So width of river is x - 49.. \[\tan(A) = \frac{x}{93}\] To upper right triangle: \[\tan(A) = \frac{x-49}{62}\] Just equate them now..
93 62 -----=---- 49+x x That's what I did
93x=3038+62x 31x=3038 x=98
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\[\frac{x}{93} = \frac{x-49}{62} \implies 62x = 93x - 93 \cdot 49 \implies 93 \times 49 = 31x \implies x = 147\] Now: \((x - 49)\) is the width of river: so: \(147 - 49 = 98\)
You are also right you can do it on your way too.. Nothing wrong in that.. But I show you what I let for each thing: |dw:1343361942290:dw|
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