An aeroplane flies from a town X on a bearing of N45°E to another town Y,a distance of 200km.it then changes course and flies to another town Z on a bearing of S60°E.if Z is directly east of X, calculate, correct to 3 significant figures,(a)the distance from X to Z (b)the distance from Y to XZ.
To solve this problem, we can use the sine rule and the cosine rule in trigonometry.
First, let's denote:
- A = the angle at X between the X-Y path and the X-Z path
- B = the angle at Y between the Y-X path and the Y-Z path
- C = the angle at Z between the Z-X path and the Z-Y path
- xy = the distance from X to Y (200km)
- xz = the distance from X to Z
- yz = the distance from Y to Z.
From the given information:
(A) A + B = N45°E - S60°E = 45° - (-60°) = 105°.
(B) We know that A + B + C = 180°, so C = 180° - 105° = 75°.
Now, let's use the sine rule to find the distance yz:
yz/sin(C) = xy/sin(B)
So, yz = xy * sin(C)/sin(B) = 200 * sin(75°)/sin(60°)
Calculating the values, we get:
yz ≈ 212.028 km.
Now, let's use the cosine rule to find the distance xz:
xz^2 = xy^2 + yz^2 - 2 * xy * yz * cos(A)
So, xz = sqrt(200^2 + (212.028)^2 - 2 * 200 * 212.028 * cos(45°))
Calculating the values, we get:
xz ≈ 192.341 km.
Therefore, the distance from X to Z (a) is approximately 192 km, and the distance from Y to the line XZ (b) can be found by using the sine rule:
sin(B)*xy = sin(A)*cq
(cq)/sin(60°)=200/sin(75°)
Calculating the values, we get:
cq ≈ 84.3 km.
So, the distance from Y to the line XZ (b) is approximately 84.3 km.