I've posted these before but still have got a good enough answer....?

Question

1) An aircraft flies from its base 200km on a bearing 162 degrees, then 350km on a bearing 260 degrees and then returns directly to the base. Calculate the length and bearing of the return journey.
I have already worked out the length, I just need the bearing.


You need to know sine and cosine rules to do this:
2) In a triangle ABC, a= Square root 28, b=6 and A =60 degrees.
a)use the cosine rule to write down a quadratic equation involving c.
b) Solve the equation to find two values of c
c) Use the Sine rule to find two values of c independently.
If you tackle this quesion, please try and lay out all the steps.

Answer 1

Nadia, this is very difficult to explain without a diagram and I hope that this is not your HW I am doing!

QUESTION 1

Step 1

Draw a triangle with starting point A. First turn at B and last turn at C. Use the bearing information and the geometry of parallel lines to find the angle at B which is 82 degrees.

Step 2

Use the cosine rule to find the distance AC

AC ^ 2 = 200^2 + 350 ^2 - 2 x 200x 350 x cos82
AC = 378.17

Step 3

Use the sine rule to find angle A

sin A / 350 = sin 82 / 378 .17

This gives angle A as 66.42

The bearing of the return journey is the angle between CA and the North line at C. This is equal to the angle between AC and the north line at A = 66.42 - 18 = 48.4

QUESTION 2

a) Do you remember that the side opposite angle A is labelled a , the side opposite angle B is labelled b and same for C

b) cos rule gives

28 = c^2 + 6^2 - 2xcx6xcos60

this simplifies to

c^2 - 6c + 8 = 0 which factorises to give you 2 values of c

c)

sin60/sqtr28 = sinB/6 so sinB = 6sin60/sqrt28 which will give you B and then C = 180 - 60 - B

followed by c/sinC = sqrt28/sin60. This will have two solutions because there will be two values of B. The one from your calculator and 180 - that.

I haven't worked all these out or given you all the steps because this looks like HW/ test paper work and it is very important that you can do all this for yourself. Please work your way through this and then try to do it without looking at my notes.

Answer 2

Oh...really, this is not complicated at all.
Take your book, and read.
Starting with the triangle(problem2) :
I think a^2=b^2+c^2-2b*c*cos A is the cosine law (look it up)
Anyway you get a 2nd degree equation
28 = 36+c^2-12*c*cosA. Again I think cos 60=1/2 (look it up)
So 28 = 36+c^2-6c that is c^2-6c+8=0 Solve this (actually c=2 is obvious) find the other solution -b+/-(b^2-4*a*c)^.5/2*a...
That's a) and b) for you.
For c) : a/sinA =b/ sinB =c/sin C. but a/sinA you already know. now plug in the b and you have sin B. My guess is that B can take 2 values (you know that any angle goes from 0 to 180), and from here you find 2 c's.

As for nr 1 I am not American and I don't know what a bearing is. It can either be an absolute angle independent of the planes flight or it can be the angle to the current direction of flight. Anyway you need to make a picture and calculate the angles of the triangle (there is just 1 triangle). If you found the lengths as you say use the cosine theorem for instance)
You can do it from here for sure. Good luck.

Answer 3

The cosine rule solves your question. the C angle?

The cosine rule is used for cases that, you have 3 sides of a tringle and you want the angle between 2 of them (In your problem, C) and in the cases that you have the length of the two sides and their between angle and want to calculate the 3rd side.

Let's solving yours. You want to find C angle. the C angle is between "a" and "b" sides, By using cosine rule we try to find the C which is between a and b. But to do this we need the third side that is c (Note that you didn't say how much is c, but i say you that this is the length of the staright way to the base in last journey)

c^2 = a^2 + b^2 - 2abcosC

we have:

cos C=(a^2 + b^2 - c^2) / 2ab

and then you will get the C!

Be succeed,
Babax.

Answer 4

for 1 I get 378,2km like above but I get 48,4 degrees

You need to figure out the distance flown North-South and then the distance flown East-West

For 200km bit, 162 degrees from North = 72 degrees from East (draw a triangle), so distance flown due South = 200 sin 72
distance flown East = 200 cos 72

For 350km bit 260 degrees from North = -10 degrees from West (draw a triangle), so distance flown South = 350 sin 10
distance flown west = 350 cos 10

Now total distance flown South = 200 sin 72 + 350 sin 10 = 251
total distance flown West = 350 cos 10 - 200 cos 72 = 282,9

Use Pythagoras, square root (251 squared + 282,9 sqared) = 378,2

then for angle draw another triangle
Angle from north = -angle from south

tan (angle)= 282,9/251

angle = 48,4 degrees

Comos method below is better but he´s screwed up the algebra

if AC/sin B = 200/sin C
then AC sin C = 200 sin B, not 200/sin B

Answer 5

Question 1
Let 1st part of journey be from A to B
This is 200 km on a bearing of 162°
Let 2nd part be from B to C
This is 350 km on a bearing of 260°
Angle ABC = 82°
AC² = 200² + 350² - 2 x 200 x 350 x cos 82°
AC = 378
A diagram is required for 2nd part:-
Start from A
D is to right of A and is a horizontal reference line.
B lies below AD such that angle DAB = 72°
AB = 200
C lies to left of A and below B
BC = 350
BF is a horizontal reference line where F lies on AC
F lies on AC and FB is horizontal
CG is horizontal line where G is to the right of C and angle BCG = 10°
Angle ABF = 72°
Angle CBF = 10°
Angle ABC = 82°
AC / sin B = 200 / sin C
AC sin C = 200 / sin B
sin C = 200 / (AC.sin B)
sin C = 200 / 378 sin 82°
C = 32.3°
Angle FCG = 42.3°
Bearing of C to A = 47.7°

Question 2
√28 / sin 60° = 6 / sin B
sin B = 6 sin 60° / √28
B = 79.1°
C = 180° - (60 + 79.1)°
C = 120°
c² = 6² + (√28)² - 2 x 6 x √28 cos 120°
c² = 64 + 31.7
c² = 95.7
c = 9.78

(now going to lie down for a wee while!)

Answer 6

For 1) I get 378.174 km at a bearing of 48.4186 degrees
sorry I reported the earlier number, it was an intermediate number from my scratch paper!

For 2)
a)
a^2=b^2+c^2-2bcCosC so,
28=36+c^2-2(6)cCos(60) Cos(60)=1/2
c^2-6c+8=0

b)
(c-4)(c-2)=0 so c=4 or 2

c)
SinA/a=SinB/b or sin(60)/sqrt(28)=sinB/6
this leads to B=79.1066 or 100.8934
since C=180-(A+B), C=40.8934 or 19.1066

c/SinC=a/SinA leads to
c=4 (for C=40.8934) or c=2 (for C=19.1066)

I hope this helps even though it took me a while!

Answer 7

Actually this problem can be solved using straight Pythagorean theorem and tangent ratios. Simply put, define your coordinate system where bearings represent the degree measures about an axis. After this you simply add the components of each of the displacement vectors (note: these are vectors meaning of course that not only do they have magnitude, they have direction) and then use a simple arctangent function to determine the return bearing.

Answer 8

oh thats next level stuff i did 3 years ago for GCSE's ! think i forgot it all the day after my exam ! but good luck with the rest of the answers hun x X x

Answer 9

Well, I too have forgotten all that sine/cosine stuff, that's why I'd draw it and use a protractor.

Answer 10

Someone that knows this stuff help this poor girl..Geeee thats what this site is for.Sorry hon I don't have a clue.