Question 1182009: A ship sails 10km on a bearing of 140 degrees and then a further 16 km on a bearing of 260 degrees (a) Calculate how far away the ship is from its starting point. (b) Calculate the bearing upon which ship must sail in order to return to its starting point
Found 2 solutions by Alan3354, math_tutor2020:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A ship sails 10km on a bearing of 140 degrees and then a further 16 km on a bearing of 260 degrees (a) Calculate how far away the ship is from its starting point. (b) Calculate the bearing upon which ship must sail in order to return to its starting point
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It's a triangle with 2 sides and the included angle given.
The included angle is 60 degs for those not familiar with bearings.
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Part (a)
Diagram:
Explanation of the diagram:
Place yourself at point A. Look directly north at point B. This is the bearing 000°. As we turn toward the east, the bearing increases. So that means 090° is directly east for instance. The bearing 140° is somewhere in the southeast quadrant.
Notice that angle BAC is 140 degrees.
Segment AC is 10 km long to indicate the ship travels 10 km along the first leg of the trip (before the 260 degree turn).
Point D is directly north of point C. It helps set up the reflex angle DCE which is 260 degrees. You start looking directly north and then turn 260° clockwise to aim at point E. Going from C to E is exactly 16 km in distance, which is the second leg of the trip.
The goal of part (a) is to find the distance from point A to point E.
From the diagram, ACD is 40 degrees (since it adds to the same side interior angle BAC = 140 to get 180; the angles are supplementary). This must mean the last missing piece is angle ACE = 60
Put another way:
(angle ACE) + (angle ACD) + (angle DCE) = (60) + (40) + (260) = 360
to represent a full 360 degree rotation
That 60 degree angle is what we'll focus on
We'll consider a triangle with
angle A = 60
side b = 10
side c = 16
Angle A is between sides b and c
We'll use the law of cosines to find the side opposite angle A, which is side 'a'
a^2 = b^2 + c^2 - 2*b*c*cos(A)
a^2 = 10^2 + 16^2 - 2*10*16*cos(60)
a^2 = 196
a = sqrt(196)
a = 14
We end up with a whole number value, which is fairly rare in these types of problems.
The distance from E to A is exactly 14 km
Answer: 14 km
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Now onto part (b)
Plot a point F anywhere directly north of point E
The bearing angle we want is the measure of angle FEA. This is the amount the ship must turn toward the east so that it can aim back home.
We'll come back to that angle later. For now, let's find angle CAE.
This is the same as finding angle A of triangle CAE.
We'll use the law of sines
sin(A)/CE = sin(C)/EA
sin(A)/16 = sin(60)/14
sin(A) = 16*sin(60)/14
sin(A) = 0.98974331861079
A = arcsin(0.98974331861079)
A = 81.786789298263
A = 81.79
The angle is approximate.
Based on this, we can say that angle CAE is roughly 81.79 degrees.
Angles BAC, CAE and EAB all combine to form a 360 degree rotation
So,
(angle BAC) + (angle CAE) + (angle EAB) = 360
(140) + (81.786789298263) + (angle EAB) = 360
221.786789298262 + (angle EAB) = 360
angle EAB = 360 - 221.786789298262
angle EAB = 138.213210701739
angle EAB = 138.21
Subtract this from 180 to find the angle FEA. This is due to the fact that angles EAB and FEA are same side interior angles (and lines FE and AB are parallel).
(angle EAB) + (angle FEA) = 180
angle FEA = 180 - (angle EAB)
angle FEA = 180 - (138.213210701739)
angle FEA = 41.786789298261
angle FEA = 41.79
The bearing is roughly 41.79 degrees
So while at point E, we will do the following to get the correct orientation
step 1) face directly north
step 2) turn approximately 41.79 degrees toward the east
This will allow us to aim at point A and help us get back home.
Answer: approximately 41.79 degrees
You may have to use the triple digit notation and say 041.79°
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