Following your data, I sketched a triange XYZ , where angle X = 45°,
angle Y = 105° and XY = 200
Find angle Z by 180-105-45 = ..
then you have a simple case of the sine law to find XZ
let the distance of Y to XZ be h
then simply:
sin45 = h/200
h = ....
An aeroplane flies from a town x on a bearing of N45°E o another town y,a distance of 200km.it then changes course and flies to another town zon a bearing of s60°e.if z is directly east of x,calculate to correct 3significant figures,
A.the distance of x to z
B.the distance from y to xz
9 answers
All angles are measured CW from +y-axis.
Given: XY = 200km[45o].
YZ [120o].
X = 90 - 45 = 45o.
Z = 120-90 = 30o.
Y = 180-45-30 = 105o.
A. XZ/sin105 = 200/sin30,
XZ = 386.4 km.
B. YZ/sin45 = 200/sin30,
YZ = ?.
Given: XY = 200km[45o].
YZ [120o].
X = 90 - 45 = 45o.
Z = 120-90 = 30o.
Y = 180-45-30 = 105o.
A. XZ/sin105 = 200/sin30,
XZ = 386.4 km.
B. YZ/sin45 = 200/sin30,
YZ = ?.
a.from triangle XYZ, <Z=180-105-45=30,than XZ/sin105=200km/sin30=386.b.when Z is directly to east of X, let represent the distance of Y to XZ with h.sin45=h/200km=141km.
Just show me the solution and answer
Please explain this question diagramatically.
Thank you
Lett h be 141km
Trignomentary
Hh each