Iam stuck with this question.Please anyone help me. An airplane is flying at an altitude of 10,000m.The pilot sees two ships A and B.Ship A is due south of P and 22.5km away(in a direct line).Likewise ship B is due East and 40.8km from P. The height is 10km. Find the distance between the two ships.
Assuming the distances are measured at sea-level, then the altitude of the plane does not affect the distance between the two ships as long as they are navigating both at sea level.
Since they form a right angled triangle with angle P, we can find the distance AB using Pythagoras theorem thus:
AB=√(PA²+PB²)
To find the distance between ship A and ship B, we can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.
Let's assign variables to the distances involved:
x = distance between ship A and ship B
y = distance between the airplane and ship A (height of the airplane)
z = distance between the airplane and ship B
Now, we can calculate the values of y and z using the given information.
The pilot sees ship A as due south, so the distance between the airplane and ship A forms the height of a right-angled triangle with the distance of 22.5 km as its base. Using the Pythagorean theorem:
y^2 = 22.5^2 - 10^2
y^2 = 506.25 - 100
y^2 = 406.25
y ≈ √406.25
y ≈ 20.16 km
Similarly, the distance between the airplane and ship B forms the height of a right-angled triangle with the distance of 40.8 km as its base. Using the Pythagorean theorem:
z^2 = 40.8^2 - 10^2
z^2 = 1664.64 - 100
z^2 = 1564.64
z ≈ √1564.64
z ≈ 39.56 km
Now, we can calculate the distance between ship A and ship B (x) by using the Pythagorean theorem again, but this time with y and z as the sides:
x^2 = y^2 + z^2
x^2 = (20.16)^2 + (39.56)^2
x^2 = 406.4256 + 1562.5936
x^2 = 1969.0192
x ≈ √1969.0192
x ≈ 44.34 km
Therefore, the distance between ship A and ship B is approximately 44.34 km.