I really need help with this. The homework question reads:
A surveyor wants to find the distance across a swamp. The bearing from A to B is N32degreesW. The surveyor walks 50 meters from A and at the point C the bearing to B is N68degreesW. Find (a) the bearing from A to C and (b) the distance from A to B.
I already got the answer for (a) that was easy. But how do I get (b)? I'm confused!!
You know the three angles and one length (from A to C) so use the sine law: [tex]\frac{c}{sin(C)}= \frac{b}{sin(B)}\) where c is the side opposite point C (i.e. the distance from A to B that you are asked to find) and b is the side opposite point B (i.e. the distance from A to C which you are given).
This one was difficult given that the solution involved understanding another geometry property that I don't believe is covered explicitly in the book, but could be argued is obvious.
Here is how to figure this one out. To start you need to redraw the diagram on a blank piece of paper as in the book and label as in the book.
Now:
1) Extend the line segment AC to go well beyond C (use a dotted line beyond C). Put a point E at the end of this line. The direction of this line is North east by the way (from A). label this additional segment as 'extension of AC' or ACE.
2) Note that the triangle ABC has a right angle at A.
3) Draw a line due 'north' from point A... straight up the paper. At the top of this line, put a point 'P1' (for parallel line number 1)
4) draw another line due north from point C ... again straight up the paper. At the top of this line, label a point 'P2'(for parallel line number 2).
5) Note that the line AC intersects both AC-P1 and AC-P2 at the same angle. So the angle made by the vertical line P1 and AC is the same angle as the angle made by the intersection of vertical line P2 and AC.
6)The problem states that Point B is at "N 32 W" of point A. That means that the right angle at point a is divided into two angles by my new line P1 (the 'north' line). The line on the left of the vertical 'north' line forms an angle of 32 degrees per the problems description. Since the total angle at A is 90 degrees, the other side of the vertical line P1 forms a 58 degree angle with the segment AC.
7. Because the two parallel lines we drew (both pointing due north through A and C respectively are both crossed by AC(and the extension to AC), then the angle formed by P2, C, and the extension (ACE) is the same as that formed by P1, A, and ACE.
8) Since the angle P1-AC is 58 degrees, the angle P2-ACE is also 58 degrees (this is the bit of knowledge not covered in the book you need to know).
9) Now look at the lines formed by P2-C and BC. These divide the line ACE into 3 component angles. We know the first angle (P2-C-E) is 58 degrees per #8 above. We know the second angle (P2-C-B) is 68 degrees because the problem description gives us that "at point C the bearing to point B is N 68 W". Given 2 of the 3 angles, we can now determine that the angle we are looking for (C) is 180 - 68 - 58 or 54 degrees.
10) Now we know angle BCA and the length of AC, so we can solve for the distance across the pond via: Tan(54 degrees) = d?/50.
Solving for D we get 68.819