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(25%) Problem 1: A fisherman spots a fish underneath the water. It appears that the fish is do = 0.33 m under the water surfaA 17% Part (b) Solve for the numerical value of L, in meters. A 17% Part (c) Express the sine of the angle Ow, in terms of @g

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- Ha do = 0.33m da: 50° nw=1-3 a nasl. tem ( 90-00) - do La do tau (90-0a) To do timog Aus L- 0.33 tem 58° : 0.520 mtre) By snells law nasinda= nusinow na sinow sinow = ha lasinda = sin oa nw as Sinow = sin so 1:32 Sino sinow = 0.652 1000 = 4

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