ZX^2 = 15^2 + 12^2 - 2*15*12cos30°
ZX = 7.565 km
Now, in triangle XYZ, we can find angle X using the law of sines
sinX/15 = sin30°/7.565
Now the bearing of Z from X is 270+(∡X-90)
Shopping centres X, Y and Z are such that Y is 12km south of X and Z is 15km from X, Z is on a bearing 330 from Y. Calculate the bearing of Z from X.
2 answers
Given: XY = 12km[180o], YZ = ___km[330o], XZ = 15km[??].
sinZ/12 = sin30/15.
Z = 24o.
330-270 = 60o
Y = 90-60 = 30o.
X = 180-24-30 = 126o.
YZ/sin126 = 15/sin30.
YZ = 24 km.
YZ = 24km[330o].
XZ = XY+YZ = 12[180o]+24[330o].
XZ = (12*sin180+24*sin330)+(12*cos180+24*cos330)i
XZ = -12 + 9i = 15km[-53o] = 15km[307o] CW.
sinZ/12 = sin30/15.
Z = 24o.
330-270 = 60o
Y = 90-60 = 30o.
X = 180-24-30 = 126o.
YZ/sin126 = 15/sin30.
YZ = 24 km.
YZ = 24km[330o].
XZ = XY+YZ = 12[180o]+24[330o].
XZ = (12*sin180+24*sin330)+(12*cos180+24*cos330)i
XZ = -12 + 9i = 15km[-53o] = 15km[307o] CW.