3 town abc are such that the distance a and b is 50km and distance between ac is 90km if the bearing of b from a is 075 and the bearing c from a is 310 find the distance bc b bearing of c from b
Draw a diagram. In triangle ABC, ∠A = 125°
So, using the law of cosines to find a (BC), we have
a^2 = 50^2 + 90^2 - 2*50*90 cos125°
BC = a = 125.55
The locations of the towns are
A = (0,0)
B = (48.3,12.94)
C = (-68.94,57.85)
So, measuring counterclockwise from the +x direction, we have
tanθ = (57.85-12.94)/(48.3+68.94) = -0.383
θ = 180-21 = 159°
So the bearing of C from B is 270+21 = 291°
Thanks you have been a good help
three towns a and c from the distance between a and b From a is 50km and the distance between a and c is 90km if the beaming of b is 75 and the bearing of c from a is 310 find
To find the distance BC and the bearing of C from B, we can use the concept of vector addition and trigonometry.
First, let's draw a diagram to visualize the given information.
```
B
\ BC
\ /
\ /
C
/ \
/ \
A \
\ \
\ \
\ /
\ /
\/
```
From the diagram, we can see that triangle ABC is a triangle with sides AB, BC, and AC. We are given the distances AB = 50 km and AC = 90 km.
Next, we need to find the distance BC. We can use the Law of Cosines to find it. The Law of Cosines states that for any triangle with sides a, b, and c, and angle C opposite to side c, the following relationship holds:
c² = a² + b² - 2ab * cos(C)
In our case, side BC is c, side AB is a, and side AC is b. Angle C, opposite to side AC, can be found using the bearing of C from A.
Let's calculate angle C:
- The bearing of C from A is 310 degrees. To convert it to an angle measured counterclockwise from the positive x-axis, we subtract it from 360 degrees: 360 - 310 = 50 degrees.
- Angle C is the bearing of C from A measured counterclockwise from the positive x-axis. So, angle C = 50 degrees.
Now, let's substitute the values into the Law of Cosines equation and solve for BC:
BC² = AB² + AC² - 2 * AB * AC * cos(C)
BC² = 50² + 90² - 2 * 50 * 90 * cos(50)
Use a calculator to compute the right side of the equation:
BC² ≈ 2750.637
Take the square root of both sides to find BC:
BC ≈ sqrt(2750.637)
BC ≈ 52.48 km (rounded to two decimal places)
Therefore, the distance BC between B and C is approximately 52.48 km.
Now, let's find the bearing of C from B. We can use the concept of trigonometry. We will use the tangent function:
tan(Bearing) = opposite/adjacent
In this case, the side opposite the angle we want to find is BC, and the side adjacent is AB.
tan(Bearing of C from B) = BC / AB
tan(Bearing of C from B) = 52.48 km / 50 km
Use a calculator to compute the right side of the equation:
tan(Bearing of C from B) ≈ 1.0496
Now, we can find the bearing of C from B by taking the inverse tangent (arctan) of the right side:
Bearing of C from B ≈ arctan(1.0496)
Use a calculator to compute the right side of the equation:
Bearing of C from B ≈ 46.66 degrees (rounded to two decimal places)
Therefore, the bearing of C from B is approximately 46.66 degrees.