Two three x and y are 100m apart and y is due east of x. The bearing of a house z from x is 145° and 225° from B. Which tree is nearer to the house and by how many metres

Answer 1

To determine which tree is closer to the house Z, we need to find the distance from both trees to the house Z.

Let's denote the distance from X to Z as a, the distance from Y to Z as b, and the distance between X and Y as d.

From the given information, we can see that a + b = 100m and a = 100 - b.

We can use the cosine rule to find the distance between Z and X and Z and Y:

cos(145) = (a^2 + d^2 - b^2) / (2ad)
cos(225) = (b^2 + d^2 - a^2) / (2bd)

Substitute a = 100 - b into the first equation:

cos(145) = ((100 - b)^2 + d^2 - b^2) / ((2 * (100 - b) * d))
cos(145) = (10000 - 200b + b^2 + d^2 - b^2) / (200d - 2bd)
cos(145) = (10000 - 200b + d^2) / (200d - 2bd)

Substitute a = 100 - b into the second equation:

cos(225) = (b^2 + d^2 - (100 - b)^2) / (2bd)
cos(225) = (b^2 + d^2 - 10000 + 200b - b^2) / (2bd)
cos(225) = (200b - 10000 + d^2) / (2bd)

Now we have two equations:

1) cos(145) = (10000 - 200b + d^2) / (200d - 2bd)
2) cos(225) = (200b - 10000 + d^2) / (2bd)

Solve these equations simultaneously to find the values of b and d. Then compare the values of a and b to determine which tree is closer to the house Z.