A salvage ship using sonar finds the angle of depressionof wreckage on the ocean floor to be 13°10'. The charts show that in this region the ocean floor is 35 meters below the surface. How far must a diver lowered from the salvage ship travel along the ocean floor to reach the wreckage? 9) A salvage ship using sonar finds the angle of depressionof wreckage on the ocean floor to be 13°10'. The chartsshow that in this region the ocean floor is 35 meters belowthe surface. How far must a diver lowered from the salvageship travel along the ocean floor to reach the wreckage?13°10'35 m

A salvage ship using sonar finds the angle of depression of wreckage on the ocean floor to be 13°10'. The charts show that in this region the ocean floor is 35 meters below the surface. To determine how far must a diver lowered from the salvage ship travel along the ocean floor to reach the wreckage.Solution: Let the labeled figure is given below. Let 'x' is the distance in meters traveled by the diver lowered from the salvage ship along the ocean floor to reach the wreckage. From the figure,DC=x Since, AB=DC⇒AB=x Consider the right ∆ABC,∠ABC=90°∠BAC=13°10'AB=xBC=35 mSince, tan∠BAC=BCABSubstituting the values we gettan13°10'=35xx=35tan13°10'Since, 13°10'=13°+10'=13°+1060° Using, 1'=160°=13°+16°=13×6+16°=796°⇒x=35tan796°Using calculator we getx=149.6147305 mTherefore, The distance traveled by the diver lowered from the salvage ship along the ocean floor to reach the wreckage is 149.6147305 meters.So, final answer=149.6147305 meters