Two house A and B are 100m apart and B is due east of A. The bearing of a church c from A is 145 and from B is 225, which house is nearer to the church and by how much
draw the triangle, label the sides, and angles.
Angle A is 55deg
Angle B is 45 deg
Angle C is 80 deg
(the ABC co-inside with the ABC above)sides a and b are opposite the angles A and B
law of sines:
100m/SinC=b/SinB solve for b
and
100m/SinC=a/SinA solve for a
To determine which house is nearer to the church, we need to find the distances from both houses to the church.
First, let's establish some notation:
- House A will be denoted as point A.
- House B will be denoted as point B.
- The church will be denoted as point C.
Given:
- The distance between houses A and B is 100m.
- The bearing of the church from house A is 145°.
- The bearing of the church from house B is 225°.
To find the distance between each house and the church, we can use the concept of trigonometry. We'll use the law of sines to solve for the distances.
Let's consider triangle ABC:
- Angle C is the bearing of the church from house A.
- Angle B is the bearing of the church from house B.
- Side a is the distance between houses A and B.
- Side b is the distance between house A and the church.
- Side c is the distance between house B and the church.
Using the law of sines, we can set up the following ratios:
b / sin(B) = a / sin(A)
c / sin(C) = a / sin(A)
To find the distances b and c, we need to know at least one of the angles B or C. Since only one angle is given, we cannot determine both distances definitively.
Therefore, we cannot determine which house is nearer to the church or by how much based solely on the given information.
To determine which house is nearer to the church and by how much, we need to calculate the distances between each house and the church.
Let's denote:
- Distance from A to the church as distance_AC
- Distance from B to the church as distance_BC
Since the bearing of the church from A is 145, we can draw a triangle with A, C (church), and a point D such that angle ACD is 35 degrees (forming a right triangle).
Using trigonometry, we can calculate the distance_AC by applying the sine function:
sin(angle_ACD) = opposite / hypotenuse
sin(35) = distance_AC / 100m
Rearranging the formula gives:
distance_AC = sin(35) * 100
distance_AC ≈ 57.24m
Similarly, let's calculate distance_BC. Since the bearing of the church from B is 225, we will consider angle BCD as 45 degrees.
Again using the sine function, we have:
sin(angle_BCD) = opposite / hypotenuse
sin(45) = distance_BC / 100m
Rearranging the formula gives:
distance_BC = sin(45) * 100
distance_BC ≈ 70.71m
Therefore, house A is nearer to the church by approximately:
distance_BC - distance_AC = 70.71m - 57.24m
≈ 13.47m
Hence, House A is nearer to the church by approximately 13.47 meters.