The bearing of a house from point A is 319°.From a point B, 317m due east of A, the bearing of the house is 288°.
Hw far is the house from A
If we label the house H, then in triangle ABH, angle A=131° and angle B=18°, so angle H is 180-131-18 = 31°
AH/sin18° = 319/sin31°
i need the the diagram ugently
I need the diagram now please
To find the distance between point A and the house, we can use the concept of trigonometry. We will form a right-angled triangle and use the information given in the problem.
Let's start by drawing a diagram:
B
*.....|
*.......|
A...H....|
*.......|
*.....|
(H represents the house, A is the starting point, and B is the point 317m due east of A)
Now, let's break down the given information:
- The bearing of the house from point A is 319°.
- The bearing of the house from point B (317m due east of A) is 288°.
First, let's find the angle between the line AB and the line AH. We can do this by subtracting the bearing of the house from both B and A:
Angle BAH = 319° - 288° = 31°
Now, we have a right-angled triangle with right angle at point H. Angle BAH is the angle opposite to the side BH, and angle ABH is the angle opposite to the side AH.
Since we have the angle BAH and the length of BH (317m), we can find the length of AH (the distance between A and H).
Using trigonometry, specifically the tangent function, we can set up the following equation:
Tan(BAH) = BH / AH
Calculating:
Tan(31°) = 317m / AH
Now, we can solve for AH:
AH = 317m / Tan(31°)
Using a calculator:
AH ≈ 580.28m
Therefore, the house is approximately 580.28 meters away from point A.