Parallax HW Problem - HELP?
Okay here's the HW problem, I would love the answer but if someone would just be able to explain to me how you're supposed to do it then that would be lovely as well. I forgot to bring my Sci stuff home over the weekend so I don't have my notes.
Here's the problem:
An archer was trying to determine the
distance of a target. He decided to use
parallax. He measured the angle from each
side of a perfectly straight, 1500 cm line.
The angle on the left side of the line was 70°
and the angle on the right side of the line was
75°.
Use parallax to determine the distance of the
target.
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Verified answer
The third angle across from the 1,500 base line is 180° - (70° + 75°) = 35°
Using the sine rule A/sina = B/sinb = C/sinc
1500cm/sin35° = 2615
Ergo:
A = 1,500 cm
B = 2615 sin70° = 2,457 cm
C = 2615 sin75° = 2,526 cm
B and C are the hypotenuses of two triangles sharing a common side. The length of that side is the distance to the target perpendicular to line A. (Beware, the angle switches team here. In the above the angle was partnered with the opposite side. Now it's opposite the shared side and the trig must follow suit.)
Distance to the target.
2457 sin75° = 2374 cm
and
2526 sin70° = 2374 cm
.
You have a triangle with a base of 15 m, and two angles.
Drop a line perpendicular to the base to the target. Now you have two right triangles.
Left triangle has base x, height y, angles 70,20,90
Right triangle has base (15–x), height y, angles 75,15,90
for the left, tan 70 = y/x
for the right, tan 75 = 7/(15–x)
two equations two unknowns
2.747477x = y
3.732051(15–x) = y
2.747477x = 3.732051(15–x)
2.747477x = 55.98076 – 3.732051x
6.479528x = 55.98076
x = 8.639635
y = 2.747477x = 23.7372 meters
check the math