Question 1182054: a ship leaves a harbour and sails at 25km/h for 3 hours on a course of N75degreesE to an island. It then sails 18km/h for 6.5hours on a course of s25degreesE to a light house. Find the distance between the harbour and the lighthouse and the bearing upon which the ship must travel to return to the harbour?
Thank you
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Draw this.
You get a triangle ABC in quadrants I and IV. A is the origin. B is 75 km away with and angle 15 degrees above the x-axis. BC and AB meet at B and the angle there is 80 degrees, between the first leg and the second leg. C is 117 km away. Let D be the point on the x-axis where AD is the line to it. This makes the triangle ABD. It has angle A of 15 degrees, B of 80 degrees, and BDA of 85 degrees. The sine of 15/x (where x is BD) as sin ABD (85 degrees)/75
this makes x 19.49 km, the length os BD.
that makes DC 97.51 km.
In the same way sin B (80 deg)/y ( AD) = sin 85/75, and AD is 74.14 km
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Now look at the triangle ADC.
We have angle ADC and two sides adjacent to the angle, AD, and DC, whose distances are known (74.14 km and 97.51 km). We also know the angle of 95 degrees.
Law of Cosines
c^2 (AC, what we want)=a^2+b^2-2ab cos C
=74.14^2+97.51^2-2(74.14*97.51)cos 95, the sqrt of which is 127.53 km
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now use the law of sines, where sin 97.5/125.53=sin C/74.14 and arc sin C=0.5791
angle C is 35.39 degrees
angle ADC is 95 degrees
so angle DAC is 49.61 degrees south of east from A so the reciprocal bearing is 229.61 degrees.
Sail at 229.61 deg for 127.53 km.
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