Given that AB = 370 m and AC = 510 m, find (i) the distance between B and C,
(ii) ACB,
(iii) the bearing of C from B,
(iv) the shortest distance from A to BC.

(i) The distance between B and C is approximately 552.9 m

(ii) ∠ACB is approximately 40.49°

(iii) The bearing of C from B is approximately 184.49°

(iv) The shortest distance from A to BC is approximately 331.155 m

Step-by-step explanation:

The given parameters are;

The length of segment AB = 370 m

The length of segment AC = 510 m

The bearing of B from A = 68°

The bearing of C from A = 144°

(i) From the bearing of B from A = 68° and the bearing of C from A = 144°, we have;

∠BAC = 144° - 68° = 76°

∠BAC = 76°

Let A represent ∠BAC, c represent segment AB, b represent segment AC, and a represent segment BC, by cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

By substituting the known values, we get;

a² = 510² + 370² - 2 × 510 × 370 × cos(76°)

a² ≈ 397000 - 91301.3234 ≈ 305698.677

BC = a ≈ √(305,698.677) ≈ 552.9

The distance between B and C, BC ≈ 552.9 m

(ii) By sine rule, we have;

a/(sin(A) = b/(sin(B)) - c/(sin(C))

Therefore;

552.9/(sin (76°)) = 370/(sin(C))

sin(C) = 370/(552.9/(sin (76°))) ≈ 0.649321

C = arcsine(0.649321) ≈ 40.49°

∠C = ∠ACB ≈ 40.49°

(iii) Angle ∠B = ∠ABC = 180° - 76° - 40.49° ≈ 63.51°

The bearing of A from B = 360° - (180° - 68°) = 248°

Therefore, the bearing of C from B ≈ 248° - 63.51° ≈ 184.49°

(iv) The shortest distance from A to BC = 370 m × sin(63.51°) ≈ 331.155 m