Town B is 50 miles north of Town A. Town C is exactly 75 miles southeast of Town A. How far is Town B from Town C?
2007-06-25 18:51:07 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You can solve this using the cosine law, but I have forgotten the formula, so I'll try an alternate solution.
The vertical component v of C's distance from A is 75*cos45.
The horizontal component h of C's distance from A is 75*sin45.
The distance d between B and C is
d = sqrt[(50 + v)^2 + h^2]
d = sqrt[(50 + 75/sqrt(2))^2 + (75/sqrt(2))^2]
d = 115.88 miles
2007-06-25 19:52:15 · answer #1 · answered by sweetwater 7 · 2⤊ 0⤋
Hello
Use the law of cosines
bc^2 = 50^2 + 75^2 -2*50*75 * cos 135
Hope This Helps!
2007-06-25 18:59:39 · answer #2 · answered by CipherMan 5 · 0⤊ 0⤋
cosine rule
x^2 = 50^2 + 75^2 - 2*50*75*cos135
= 13428.3
x = 115.88 miles
2007-06-25 18:54:38 · answer #3 · answered by Dr D 7 · 0⤊ 0⤋
Town A = (0, 0)
Town B = (0, 50)
Town C = (75 / sqrt(2), -75 / sqrt(2))
distance = sqrt[ (75 / sqrt(2) - 0)^2 + (-75 / sqrt(2) - 50)^2 ]
distance = 115.9
Answer: 115.9 miles
2007-06-25 19:17:59 · answer #4 · answered by mathjoe 3 · 0⤊ 0⤋
Ask what number miles is a million/5 of 20 miles (a million inch), and to get the answer to that, divide 20 miles via 5, equals 4 miles. meaning 20 miles is 4 miles taken 5 instances. for this reason, a 4 miles stretch is a million/5 of an inch on the map. upload a hundred and twenty miles (20 miles x 6 in.) to 4 provides 124 miles.
2017-01-01 05:40:14 · answer #5 · answered by bashford 3 · 0⤊ 0⤋
using the cosine rule which is :
a^2 = b^2 + c^2 -2bccosA
a^2 = (50)^2 + 75^2 - 2*50*75*cos135
a = 115.88
2007-06-25 20:08:09 · answer #6 · answered by mistu 2 · 0⤊ 0⤋
i certainly would not use pythagorean theorem on this one. but it certainly stomped me for lack of an angle.
2007-06-25 19:36:02 · answer #7 · answered by Ω allan y 6 · 0⤊ 0⤋
90.14 MILES
Your right ZEPOL my boo boo.
2007-06-25 18:59:59 · answer #8 · answered by PC 7 · 0⤊ 1⤋