Two boat leave a port at the same time.the first travels at 15km/hr on a bearing 135° while the second travel at 20km/hr on the bearing of063° if after 2hours,the second boat is directly north of the first boat,calculate their distance apart(showing the diagram)
2 answers
Please i need a picture of the diagram to that question
You have contradictory information
From the first part of your description, we can sketch the diagram
and use the cosine law to find the distance apart
After 2 hours, the first boat went 40 km and the second went 30 km
D^2 = 40^2 +30^2 - 2(40)(30)cos 72°
D = 41.93 km
let the angle formed at the position of the 63° direction boat be A
sinA/30 = sin72/41.93
sinA = .6804...
angle A = 42.88°
but if this boat is directly north of the other boat, that angle should be 63°
check your question and/or your typing
From the first part of your description, we can sketch the diagram
and use the cosine law to find the distance apart
After 2 hours, the first boat went 40 km and the second went 30 km
D^2 = 40^2 +30^2 - 2(40)(30)cos 72°
D = 41.93 km
let the angle formed at the position of the 63° direction boat be A
sinA/30 = sin72/41.93
sinA = .6804...
angle A = 42.88°
but if this boat is directly north of the other boat, that angle should be 63°
check your question and/or your typing
1 answer