A boat leaves a port at 10AM traveling at a speed of 30 mi/h in the direction N50°E. Another boat leaves the same port at the same time traveling at a speed of 26 mi/h in the direction S70°E. How long does it take for the boats to be 75 miles a part?
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Let the time for that to happen be t hours after 10:00 am
As usual make a diagram,
the distance travelled by 1st boat is 30t miles
the distance travelled by 2nd boat is 26t miles
the angle between their paths is 60° , (look at your diagram)
by the cosine law
75^2 = (30t)^2 + (26t^2) - 2(30t)(26t)cos60°
5625 = 900t^2 + 676t^2 - 780t^2
solve for t, add it to 10:00 am
let me know what you get
Looked at the wording again:
It said "How long does it take for the boats to be 75 miles a part?"
so all we need it the value of t
the 10:00 am is just extra noise
So first, use the bearings to find the difference between the angles the ships move at. So that would be (90-50)+(90-70) which is 60 degrees.
Then, use the law of cosines to find the missing side, with 26 and 30 as the two sides:
26^2 + 30^2 - 2(26)(30)(cos60) = c^2
Once you solve, you should get that side C is approximately 2 *sqrt(199)
Then, since you want to know how long it will take to get to 75 miles, you divide 75 by 2 *sqrt(199) and you should get about 2.6583 hours