Question

a surveyor, walking on a level ground at 10 km/hr in the eastward direction, sighted the foot of a tall narra tree in the direction E 12 degrees North. after 30mins, the same tree now bears East 48 degrees North . how far was he from the tree during the first sighting?

Answer 1

That's a fast walk!

Answer 2

could you help me to answer this question ?

Answer 3

Yes, working on it now...

Answer 4

Ok, I think the situation looks like this:

Answer 5

yah that is the exact illustration here in my book .

Answer 6

You have angle-side-angle information about a triangle; you can easily find the third angle, then use a combination of law-of-sines and law-of-cosines to solve.

Answer 7

the answer is 36 degrees?

Answer 8

That's your third angle, yes. Do you know the law of sines and law of cosines?

Answer 9

I think law of sines is all you need for this one.
\[\frac{a}{\sin(A)}=\frac{b}{\sin(B)}\]

Answer 10

I really can't remember how to use the law of sine and low of cosine.

Answer 11

what about the law of cosine?

Answer 12

Law of cosines is better when you have a side-angle-side situation. You can still use it for redundancy to check you answer.

Answer 13

so the answer is 7.01 km?

Answer 14

Yep, that looks good to me.

Answer 15

thank you I will post another question tomorrow I hope you find time to answer my questions.

Answer 16

Just in case, I'm going to check it using law of cosines. Sometimes with obtuse triangles, sines can give ambiguous answers.

Answer 17

@exeal
Yeah, ran into some obtuse triangle trouble. Law of cosines gives a different answer.
Both equations agree on d=6.32km. 7.01 is erroneous.

Answer 18

so I will use the law of cosine. Thank you again

Answer 19

I always use both to be on the safe side.

Answer 20

You're welcome.