A girl starts from Point A and walks to 285m to B on a bearing of 078°. She then walks due south to a point C which is 307m from A. What is the bearing of A from C and what is IBCI

bearing of A from C is - 65.24°

the distance |BC| is 187.84 m

Step-by-step explanation:

given data

girl walks AB = 285 m (side c)

bearing angle B = 78°

girl walks AC = 307 m (side a)

solution

we use here the Cosine Law for getting side b that is

ac² = ab² + bc² - 2 × ab × cos(B)     1

307² = 285² + x²  - 2 × 285 cos(78)

x = 187.84 m

and

now we get here angle θ , the bearing from A to C get by law of sines

sin (θ) =

sin (θ) =  0.5985  

θ = 36.76°

and as we get here angle BAC that is

angle BCA = 180 - ( 36.76° + 78° )

angle BCA = 65.24°

and here  negative bearing of A from C so - 65.24°

Step-by-step explanation:

3x + 18 = 147°

step-by-step explanation:

3x + 18 = 147°, because the value of the central angle is equal to the value of the arc formed by it.

i hope i you.

it would rotate about 180 degrees.