A ship is anchored off a long straight shoreline that runs north and south. From two observation points 15 mi apart on shore, the bearings of the ship are N 31 degrees E and S 53 degrees E. What is the shortest distance from the ship to the shore?

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Answer 1 · 2018-07-13T11:28:09

$6.2$ $color(magenta)(6.2$ $miles to the nearest 1 decimal place$ $"miles to the nearest 1 decimal place"$

$:."Let the triangle with the ship and shorline be ABC and let the shortest distance be AD"$

$:.angle ADC and angle ADB=90^@$

$:.In triangle ABD angle "BAD"= 180^@-(90^@+31^@)=59^@$

$:.In triangle ACD angle "CAD"= 180^@-(90^@+53^@)=37^@$

$:.angle BAC=59^@+37^@=96^@$

Sine rule

$:.(AB)/sinC=(BC)/sinA$

$:.AB=(BC*sinC)/sin A$

$:.AB=(15*sin 53)/sin 96^@$

$:.AB=12.046mi$

$:.AD=sin31^@*12.046$

$:.color(magenta)(=6.204mi="shortest distance"$

check:-

$:.(AC)/sinB=(AB)/sinC$

$:.(AC)/(sin31^@)=(12.046*sin31)/sin53^@$

$:.AC=7.768mi$

$:.color(magenta)(AD=cos37^@*7.768=6.204mi$