Bearing questions - plz help

• May 23rd 2010, 03:50 AM
dee2020
Bearing questions - plz help(SOLVED)
1.A hiker walks due north from A and after 8km reaches B. She then walks a further 8km on a bearing of 120degrees to C. Work out a the distance from A to C and b the bearing of C from A.

2.A helicopter flies on a bearing of 200degrees from A to B, where AB =70km. It then flies on a bearing of 150deg from B to C, where C is due south of A. Work out the distance of C from A.

3.Two radar stations A and B are 16km apart and A is due north of B . A ship is known to be on a bearing of 150deg from A and 10km from B. Show that this information gives two positions for the ship, and calculate the distance between these two positions.

Any help or hint on how to solve will be much appreciated.

dee
• May 23rd 2010, 04:16 AM
skeeter
Quote:

Originally Posted by dee2020
1.A hiker walks due north from A and after 8km reaches B. She then walks a further 8km on a bearing of 120degrees to C. Work out a the distance from A to C and b the bearing of C from A.

make a sketch ... ABC is equilateral

2.A helicopter flies on a bearing of 200degrees from A to B, where AB =70km. It then flies on a bearing of 150deg from B to C, where C is due south of A. Work out the distance of C from A.

make a sketch, determine the measure of angle B, and then use the law of cosines.

3.Two radar stations A and B are 16km apart and A is due north of B . A ship is known to be on a bearing of 150deg from A and 10km from B. Show that this information gives two positions for the ship, and calculate the distance between these two positions.

make a sketch ... ambiguous case of the law of sines.

...
• May 23rd 2010, 05:12 AM
HallsofIvy
Quote:

Originally Posted by dee2020
1.A hiker walks due north from A and after 8km reaches B. She then walks a further 8km on a bearing of 120degrees to C. Work out a the distance from A to C and b the bearing of C from A.

180- 120= 60 so the angle between the "due south" line from B to A to the line from B to C is 60 degrees. You have a triangle with two sides of length 8 and angle between them of 60 degrees. You could use the cosine law: [tex]c^2= a^2+ b^2- 2abcos(C)= 8^2+ 8^2- 2(8)(8)cos(60). However, here, you have a triangle with two sides of equal length and a 60 degree angle. What kind of triangle does that remind you of?

Quote:

2.A helicopter flies on a bearing of 200degrees from A to B, where AB =70km. It then flies on a bearing of 150deg from B to C, where C is due south of A. Work out the distance of C from A.
Draw a picture, drawing the triangle ABC. 200- 180= 20 so the angle between line AB and "due south" line AC is 20 degrees. That is the angle at A in the triangle ABC. 200- 150= 50 degrees and that is the angle outside the triangle at B. The angle inside the triangle is 180- 70= 110 degrees. Of course, the third angle is 180- 20- 110= 50 degrees. Use the sine law $\displaystyle \frac{sin(A)}{a}= \frac{sin(B)}{b}= \frac{sin(C)}{c}$ $\displaystyle \frac{sin(20)}{a}= \frac{sin(110)}{b}= \frac{sin(50)}{70}$ to solve for a and b. b is the "distance of C from A".

Quote:

3.Two radar stations A and B are 16km apart and A is due north of B . A ship is known to be on a bearing of 150deg from A and 10km from B. Show that this information gives two positions for the ship, and calculate the distance between these two positions.
Draw a picture with A directly above B, length of AB= 16. The ship, C, will be to the nw of A. If you draw lines due east from C and due north from A, you have a right triangle with angle 150- 90= 60 at C so the angle in that right triangle at A is 90- 60= 30 degrees and the angle at A in triangle ABC is 180- 30= 150 degrees (Huh! We could have gotten that by treating AC as a transversal on two vertical, parallel lines). We now have a triangle with angle of 150 degrees at A and lengths AB= 16, BC= 10. Those two sides do NOT have angle A between them but we can use the cosine law: $\displaystyle c^2= a^2+ b^2- 2abcos(C)$ so $\displaystyle 10^2= a^2+ 16^2- 2(16)a cos(150)$. That gives a quadratic equation to solve for a. Since it is quadratic you can expect it to have two solutions.

Quote:

Any help or hint on how to solve will be much appreciated.

dee
I stopped for a telephone call and skeeter got in ahead of me!
• May 23rd 2010, 05:12 AM
Soroban
Hello, Dee!

Quote:

1.A hiker walks due north from $\displaystyle A$ and after 8km reaches $\displaystyle B.$
She then walks a further 8km on a bearing of 120° to $\displaystyle C.$
Find: (a) the distance from $\displaystyle A$ to $\displaystyle C$, and (b) the bearing of $\displaystyle C$ from $\displaystyle A.$

Code:

      :       : 120°     B o       |  *  8       | 60° *       |        *     8 |          o C       |        *       |    *       |  *     A o

Triangle $\displaystyle ABC$ is equilateral.
We have all the information we need.

Quote:

2. A helicopter flies on a bearing of 200° from A to B, where AB = 70km.
It then flies on a bearing of 150° from B to C, where C is due south of A.
Find the distance of C from A.

Code:

                        P                         :                         :                         o A                     *  |       Q      70  * 20° |       :        *        |       : 20° *          |       :  *              |     B o 130°            | x       :  *              |       : 30° *          |       :        *        |       R          * 30° |                     *  |                         o C

Major angle $\displaystyle PAB = 200^o \quad\Rightarrow\quad \angle BAC = 20^o = \angle QBA$

Angle $\displaystyle QBC = 150^o \quad\Rightarrow\quad \angle RBC = 30^o = \angle BCA$

Hence: .$\displaystyle \angle ABC = 130^o$

Let $\displaystyle x = AC.$

Law of Sines: .$\displaystyle \frac{x}{\sin130^o} \:=\:\frac{70}{\sin30^o}$

• May 31st 2010, 02:48 AM
dee2020
@skeeter : Thanks a lot skeeter
@hallsofIvy : ur so funny,,,thanks so much
@soroban: Thanks very much

Thanks all again.