Asked by Anonymous

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.

Answers

Answered by Reiny
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
Answered by Esther
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.
Answered by joshuana
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
Answered by Anonymous
no sketch
Answered by Ezekiel
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.
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