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

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )