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
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.
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 = ?.
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 = ....
Please explain this question diagramatically.
Just show me the solution and answer
Trignomentary
Lett h be 141km
Thank you
Hh each
To solve this problem, we can use trigonometry and the concept of bearings to find the distances.
A. The distance from town x to town z:
Since the plane flies on a bearing of N45°E from town x to town y, we can visualize this as a right-angled triangle. The bearing angle of N45°E can be split into two angles, N45° and E45°.
Using trigonometry, we can determine the distance traveled in the north direction (N45°) and the distance traveled in the east direction (E45°).
The distance traveled in the north direction can be calculated using the sine function:
Distance north = 200 km * sin(45°) ≈ 141.42 km (rounded to 3 significant figures)
The distance traveled in the east direction can be calculated using the cosine function:
Distance east = 200 km * cos(45°) ≈ 141.42 km (rounded to 3 significant figures)
Now, to find the distance from town x to town z, we have the following right-angled triangle:
|
y | x
______|______
Using the Pythagorean theorem, we can find the hypotenuse (the distance from x to z):
Distance xz = √((Distance north)^2 + (Distance east)^2)
= √((141.42 km)^2 + (141.42 km)^2)
≈ √(20000 km^2)
≈ 141.42 km (rounded to 3 significant figures)
Therefore, the distance from town x to town z is approximately 141.42 km.
B. The distance from town y to town xz:
Since we have already determined the distance from town x to town z, we can now find the distance from town y to town xz.
The bearing angle from town y to town xz is s60°e. This means that the plane has to fly on a bearing of 180° + 60° = 240° from town y to town xz.
Using the concept of bearings, the plane must travel in the opposite direction to reach town xz from town y.
Therefore, the distance from town y to town xz is the same as the distance from town x to town z, but in the opposite direction.
So, the distance from town y to town xz is also approximately 141.42 km.
To summarize:
A. The distance from x to z is approximately 141.42 km.
B. The distance from y to xz is also approximately 141.42 km.