a plane flies from A to B , a distance of 120 kilometres on a bearing of 035 degrees. he then changes direction and flies to a further distance of 200 kilometres in a bearing of i25 degrees. find the distance AC to the nearest kilometre
I will assume the 2nd direction is 125°
The angle between the two directions is 90°
so AC = √(120^2 + 200^2)
= 233.238 km
or
Suppose we did not see that 90° ,we could simply use vectors
AC = (120cos35,120sin35) + (200cos125,200sin125)
= (98.298.., 68.829..) + (-114.715.. , 163.830..)
= (--16.417.. , 232.659..)
whose magnitude = √( (-16.417)^2 + 232.959^2)
= appr 233.238 km
To find the distance AC, we need to find the length of leg AC in a right-angled triangle formed by points A, C, and B.
First, let's calculate the horizontal and vertical distances traveled by the plane from A to B.
Horizontal Distance (AB):
AB can be calculated using the formula: AB = cos(degree) * distance
AB = cos(035) * 120 km
AB ≈ 0.819 * 120 km
AB ≈ 98.28 km (rounded to two decimal places)
Vertical Distance (BC):
BC can be calculated using the formula: BC = sin(degree) * distance
BC = sin(035) * 120 km
BC ≈ 0.574 * 120 km
BC ≈ 68.88 km (rounded to two decimal places)
Now, we need to calculate the horizontal and vertical distances traveled by the plane from B to C.
Horizontal Distance (BC):
BC can be calculated using the formula: BC = cos(degree) * distance
BC = cos(125) * 200 km
BC ≈ -0.573 * 200 km (negative sign indicates movement in the opposite direction)
BC ≈ -114.6 km (rounded to one decimal place)
Vertical Distance (AC):
Since point C is directly below point A, the vertical distance AC will be equal to the vertical distance BC.
AC ≈ 68.88 km (rounded to two decimal places)
Therefore, the distance AC is approximately 69 kilometers to the nearest kilometer.