A man sets out from an oasis to walk acorss the desert to a town which is 55 km due north. Unfortuanately, there is a mountain that he must walk around the reach the town. On the first day walks 20 km on a bearing of 340 degrees. On the second day he walks another 20 km on a bearing of 20 degrees.

Find

The shortest remaining distance to the town
And
The bearing on which the man must walk to reach the town

The angle where he turns is 140 degrees(see why?).

Use the law of cosines.

c=\sqrt{20^{2}+20^{2}-2(20)(20)cos(140)}

He ends up directly south of the town. Just subtract the result from 55 to find the distance he has yet to go.

No sir I don't see how you got 140 degrees... im sorry
Can you please explain me

And sir in my math text book
The answer is 17 km and the bearing is 000 degrees north
The answer you gave doesn't match the answer I have in the book sir

I did not give the answer.

What did you get using the law of cosines?

Subtract that from 55 and you should get approximately 17 rounded down.

The 140 comes from looking at the diagram. Turn around and face the other direction. Do a 180. 340-180=160. The other angle is 20, so 160-20=140.

That is the angle at his first turn

The answer I got when using the cosine rule
Was 37. 59 sir
And when subtracted form 55 it gives an answer below 0 sir.

Think for a second:rolleyes:. If you subtract 37.59 from 55, how can the answer be less than 0?

Omg sir
I'm so sorry
Yesterday I wasn't very focused
But thank you so much sir
I was doing 37.59- 50
How foolish of me

But sir one more question please
How do I find the bearing again on which the man must walk to reach the town?

On your diagram Emporer! You'll see that the man will be directly at the south of the town, like galactus said some post earlier:


He ends up directly south of the town

Therefore, the man has to go north. But what is the bearing of north? That, you'll see as 0 degrees.