Question

Challenge fr y'all. solve this to get 50 points and brainliest coming up if you get the right answer

Answer 1

Answer:

5 Hours and 12 k/m

Hope this helps :)

Answer 2

Answer:

• See below

Step-by-step explanation:

a)

i)

Use the law of cosines:

• $AB=\sqrt{PA^2+PB^2-2*PA*PB*cosP}$
• $AB=\sqrt{60^2+100^2-2*60*100*cos80} =\sqrt{11516.22} =107.31$

ii)

Use the law of sines:

• sin ∠A / PB = sin ∠P / AB
• sin ∠A / 100 = sin 80 / 107.31
• sin ∠A = 100 sin 80 deg / 107.31
• sin ∠A = 0.92
• ∠A = arcsin 0.92
• ∠A = 67° (rounded)

iii)

Bearing of B from A:

• 90° + (67° - 30°) =
• 127°

b)

PB is longer distance. Time to reach B:

• t = d/s
• t = 100/20 = 5 hours

The speed s of the slower ship:

• s = d/t
• s = 60/5 = 12 km/h