Asked by ka
Angela traveled a distance of 42km from A, on a bearing of 080 degree to B. At B she then traveled a further distance of 20km to C on the bearing of 160 degree. Calculate a) distance/AC/
b) the bearing of A from C.
b) the bearing of A from C.
Answers
Answered by
@ben/rose/bose/ka
See ken's post, and come on back with some of your ideas.
Always start by drawing the diagram
recall how to convert polar form to x-y coordinates for adding up distances
and study up on law of cosines
Always start by drawing the diagram
recall how to convert polar form to x-y coordinates for adding up distances
and study up on law of cosines
Answered by
Damon
...... and please note that you take a bearing on something to find its direction from you.
HOWEVER if you sail for it, you travel on a HEADING not a BEARING.
HOWEVER if you sail for it, you travel on a HEADING not a BEARING.
Answered by
henry2,
All angles are measured CW from +y-axis.
AC = AB + BC = 42km[80o] + 20km[160o].
X = 42*sin80 + 20*sin160 = 48.2 km.
Y = 42*Cos80 + 20*Cos160 = -11.5 km.
a. AC = sqrt ( X^2+Y^2).
b. TanA = X/Y.
AC = AB + BC = 42km[80o] + 20km[160o].
X = 42*sin80 + 20*sin160 = 48.2 km.
Y = 42*Cos80 + 20*Cos160 = -11.5 km.
a. AC = sqrt ( X^2+Y^2).
b. TanA = X/Y.
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