C2 Bearings

While revising for C2 I stumbled upon a question which involves bearings.

Here's the question:

A lifeboat station is at point A. A distress call is received and the lifeboat travels 15km on a bearing of 030° to point B. A second call is received and the lifeboat then travels 27km on a bearing of 110° to arrive at point C. The lifeboat then travels back to the station at A.

(i) Show that angle ABC is 100°

(ii) Find the distance that the lifeboat has to travel to get from C back to A

(iii) Find the bearing on which the lifeboat has to travel to get from C to A.

I proved part one:

360-(150+110)=100

I solved part (ii) with the cos rule and got 33.08 for the angle

I have trouble with point (iii) cause I have no idea how to find out the bearing there.
I found the angle C using the sin rule

sin(100)/33.08=sinC/15

Angle C=26.5 degrees

Now how do I go on from here to find the bearing from C to A?

Reply 1

Indeterminate

3

Original post by Blackfyre

While revising for C2 I stumbled upon a question which involves bearings.

Here's the question:

A lifeboat station is at point A. A distress call is received and the lifeboat travels 15km on a bearing of 030° to point B. A second call is received and the lifeboat then travels 27km on a bearing of 110° to arrive at point C. The lifeboat then travels back to the station at A.

(i) Show that angle ABC is 100°

(ii) Find the distance that the lifeboat has to travel to get from C back to A

(iii) Find the bearing on which the lifeboat has to travel to get from C to A.

I proved part one:

360-(150+110)=100

I solved part (ii) with the cos rule and got 33.08 for the angle

I have trouble with point (iii) cause I have no idea how to find out the bearing there.
I found the angle C using the sin rule

sin(100)/33.08=sinC/15

Angle C=26.5 degrees

Now how do I go on from here to find the bearing from C to A?

Use your sketch, noting where bearings are measured from :smile:

Reply 2

0x2a

2

Use the sine rule to get angle BCA, and use the rule that interior angles add up to 180 degrees (At B & C).

Now add those 2 values together and take it away from 360 degrees and you'll have the bearing from C to A.

Reply 3

Blackfyre

OP

10

Original post by 0x2a

Use the sine rule to get angle BCA, and use the rule that interior angles add up to 180 degrees (At B & C).

Now add those 2 values together and take it away from 360 degrees and you'll have the bearing from C to A.

I noted in the thread that I calculated BCA, it's 26.5

How do I use the rule that all angles add up to 180 in this situation?

Reply 4

0x2a

2

Original post by Blackfyre

I noted in the thread that I calculated BCA, it's 26.5

How do I use the rule that all angles add up to 180 in this situation?

The 2 angles between the North-line and line segment B and C are interior angles. You already know the angle between the North-line and point B, so you can work out the one at point C.

Reply 5

Blackfyre

OP

10

Ah got it, so it's 180-110=70

360-(70+26.5)=263.5

Cheers.

C2 Bearings

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