two trees x and y are 100m a part and y is due east of x. The bearing of a house from x is 145° and 225° from B which tree is nearer to the house and by how many meters
Let's label the house as point H. First, let's understand the bearings given:
- The bearing of the house from tree X is 145°, which means the angle XHY is 180° - 145° = 35°.
- Similarly, the bearing of the house from tree Y is 225°, which means the angle YHX is 225° - 180° = 45°.
Now, we have a triangle with vertices X, Y, and H. We know that angle XHY is 35°, angle YHX is 45°, and angle XYH is 180° - (35° + 45°) = 100°. We also know that the distance between the trees is 100m (XY = 100m).
Let's apply the sine rule on this triangle:
(XH/sin(YH)) = (YH/sin(XH)) = (XY/sin(XYH))
Now we can solve this system for XH and YH:
XH/sin(100°) = YH/sin(35°) = 100m/sin(45°)
We know sin(35°) = 0.5736, sin(45°) = 0.7071, and sin(100°) = 0.9848.
To find XH:
XH = 100m * sin(45°) / sin(100°)
XH = 100 * 0.7071 / 0.9848
XH ≈ 71.7m
To find YH:
YH = 100m * sin(35°) / sin(100°)
YH = 100 * 0.5736 / 0.9848
YH ≈ 58.3m
So, tree Y is nearer to the house by (71.7 - 58.3) m = 13.4 meters.