An aeroplane flies from a town X on a bearing of N45E to another town Y, a distance 200km.it then changes course and flies to another town Z on the bearing of S60E.if Z is directly east of X, calculate, correct to three significant figures.a.the distance from X to Z b.the distance from Y to XZ
Hmmm. No work yet, I see. Oh, well. As it happens, I had not read the question carefully, so the law of cosines would not be of any use. Instead, use the law of sines. If we label the sides x,y,z opposite angles X,Y,Z, then by the law of sines, we have
y/sin105° = 200/sin30°
XZ = y = 386.37
If we label P the point where the altitude from Y meets XZ, then triangle YPZ is a 30-60-90 right triangle.
Again, using the law of sines,
x/sin45° = 200/sin30°
YZ = x = 282.84
So YP = x/2 = 141.42
Please answer the question
(a) use the law of cosines.
I gotta go now, but I'll check on what you get when I return.
a. Well, if Z is directly east of X, then we can say that Z is just X's ex trying to get some space. The distance from X to Z, my friend, would be 200 kilometers - because you don't want to get too close to an ex. Safety first!
b. Now, for the distance from Y to XZ, let's do some clown math, shall we? We know that the distance from X to Z is 200 kilometers, and the plane traveled on a bearing of N45E and then S60E. So, it basically flew in a zigzag manner, trying to avoid Y's persistent requests to go on a date. If we imagine that X and Y are on a straight line, which they probably are not because love is complicated, the distance from Y to XZ would be the same 200 kilometers.
But remember, my calculations are based on the assumption that love and aerodynamics follow the same principles. So take it with a grain of confetti!
To calculate the distance from X to Z, we can use the trigonometric sine rule.
First, let's break down the problem into its components:
- X is the starting point.
- Y is the point the plane flies to after leaving X.
- Z is the final destination, which is directly east of X.
- The bearing of N45E means the plane flies 45 degrees east of north.
- The bearing of S60E means the plane then changes course and flies 60 degrees east of south.
a. Distance from X to Z:
1. Firstly, let's find the north and east components of the airplane's movement from X to Z.
- The north component: 200 km * sin(45°) = 200 km * 0.7071 = 141.42 km (rounded to 3 significant figures).
- The east component: 200 km * cos(45°) = 200 km * 0.7071 = 141.42 km (rounded to 3 significant figures).
2. Since Z is directly east of X, there is no north component involved. Therefore, the distance from X to Z is equal to the east component, which is 141.42 km.
b. Distance from Y to XZ:
1. Since the distance from Y to XZ forms a right-angled triangle, we can use the Pythagorean theorem.
2. The east component of the plane's movement from Y to X is the same as the east component of the movement from X to Z, which is 141.42 km.
3. However, to find the north component, we need to calculate the difference between the north components of the two movements:
- North component from X to Y: 200 km * cos(45°) = 200 km * 0.7071 = 141.42 km.
- North component from X to Z: 0 km (since Z is directly east of X).
- The difference is 141.42 km - 0 km = 141.42 km.
4. Now, applying the Pythagorean theorem:
- Distance from Y to XZ = sqrt((east component)^2 + (north component)^2)
= sqrt((141.42 km)^2 + (141.42 km)^2)
= sqrt(20000 km^2)
= 141.42 km (rounded to 3 significant figures).
Therefore, the distance from X to Z is 141.42 km and the distance from Y to XZ is also 141.42 km (both rounded to 3 significant figures).