Two boats leave a port at the sametime. The first travels at 15km/hr on a bearing of 135 degree. While the second travels at 2km/hr on a bearing of 63 degree . If after 2hours the second boat is directly North of the first boat, calculate their distance apart.

Answer 1

So the angle between them is 72°
They both travel for 2 hours, so the 1st went 30 km, and the 2nd went 4 km

This is just a cosine law question ....

x^2 = 30^2 + 4^2 - 2(4)(30)cos72
= 841.83..
x = appr 29.01 km

or, using the fact that the 2nd boat was directly north of the 1st, and noting parallel lines
4/sin45 = x/sin72
xsin45 = 4sin72
x = 4sin72/sin45 = 5.37 km

which is inconsistent with my other answer.
I think you have contradictory information.
Make a sketch and you will see that 2nd boat cannot be directly north of the 1st.

Check your typing or your question itself

Answer 2

The distance should be gotten so as to know the bearing of the two boats. Since an average speed equals distance over time, therefore distance will be a.v(average speed)*time.
Distance=15*2
=30km for the first boat
And for the second boat,
Distance =20*2=40km
The angle between them is 72°.
Using cosine rule,
a^2=b^2+c^2-2bccosA
a^2=40^2+30^2-2*40*30cos72
a^2=1600+900-2400(0.309)
a^2=2500-741.6
a^2=1758.4
a=√1758.4
a=41.93~42km.

Answer 3

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 A BEARING NORTH OF THE FIRST BOAT CALCULATE THEIR DISTANCE A PART

Answer 4

no sketch

Answer 5

Why can't we use some rule for this question?

If we take the angle from at the port and the angle formed at the BOAT A, I believe we can use SINE RULE.