Answer / poojag

Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100 m, ACB = 30° and ADB =45

CD = (AC + AD) = (100√3 + 100) m
= 100(√3 + 1)
= (100 x 2.73) m
= 273 m.

Answer / 100r@bh

distance between the two boats
Let AB be the light house and the two boats be at C and D
AB = 100 m
tan30° = BC/AB | BC = tan(30) * 100 = 57.7350
BC = 57.7350 m
tan45° = BD/AB | BD = tan(45) * 100 = 100
BD = 100 m
Therefore, the distance between the two boats is = BC+BD
= 57.7350+ 100
= 157.73 m

IF you find Both 2 Boat Distance from light house
AB2 + BC2 = AC2 || Square root(1002 + 57.73502 ) = 115.47 || AC = 115.47
AB2 + BD2 = AD2 || Square root(1002 + 1002 ) = 115.47 || AD = 141.42

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