Asked by Joy
A road starts at a college and goes due north for 2000m, it then goes 2000m on a bearing 040 degree and ends at a market. How far is the market from the college
Answers
Answered by
Reiny
I will assume you made a sketch
by the cosine law:
x^2 = 2^2 + 2^2 - 2(2)(2)cos100°
solve for x
multiply the answer by 1000, since I scaled it back
careful with that last term, remember cos100° is negative.
by the cosine law:
x^2 = 2^2 + 2^2 - 2(2)(2)cos100°
solve for x
multiply the answer by 1000, since I scaled it back
careful with that last term, remember cos100° is negative.
Answered by
Reiny
arghhhh!
my angle should have been 140°, but I am sure you had that
figured out.
my angle should have been 140°, but I am sure you had that
figured out.
Answered by
henry2,
All angles are measured CCW from +x-axis.
d = 2km[90o] + 2km[140o].
X = 2*Cos90 + 2*Cos140 = 0 - 1.53 km. = -1.53 km.
Y = 2*sin90 + 2*sin140 = 2 + 1.29 = 3.29 km.
d = -1.53 + 3.29i = 3.63km[-65o] = 3.63km[65o] N. of W. = 3.63km[115o].
3.63 km = 3630 m.
d = 2km[90o] + 2km[140o].
X = 2*Cos90 + 2*Cos140 = 0 - 1.53 km. = -1.53 km.
Y = 2*sin90 + 2*sin140 = 2 + 1.29 = 3.29 km.
d = -1.53 + 3.29i = 3.63km[-65o] = 3.63km[65o] N. of W. = 3.63km[115o].
3.63 km = 3630 m.
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