A , B and C are three shops such that B is 600 m due East of A. The bearing of C from A is 144∘ . The bearing of C from B is 240∘ . Find the distance AC, correct to 1 decimal place.Please give 1 answer.302.0 m301.6 m301.0 m301.7 m

To solve this problem, we can use the law of sines in trigonometry.

First, we need to find the angle ∠ACB. Since we know that ∠CAB is 144° and ∠CB is 240°, we can find ∠ACB by subtracting 180° from the sum of ∠CAB and ∠CB.

∠ACB = (144° + 240°) - 180° = 204°

Next, we can use the law of sines to find the length of AC. The law of sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle.

So, we have:

AC/sin(∠CAB) = AB/sin(∠ACB)

Substituting the given values:

AC/sin(144°) = 600m/sin(204°)

Solving for AC gives:

AC = (600m * sin(144°))/sin(204°)

Using a calculator to compute the sines and perform the division gives:

AC ≈ 301.6m

So, the correct answer is 301.6 m.