It should be easy to see that the angle between them is 110 degrees.
Now it becomes a case for the cosine law.
Be careful with the sign of the -2(2)(5)cos 110 term in the equation.
Let me know what you get.
FROM A POINT ON THE EDGE OF THE SEA,ONE SHIP IS 5KM AWAY ON A BEARING SOUTH 50 DEGREES EAST AND ANOTHER IS 2KM AWAY ON A BEARING SOUTH 60 DEGREES WEST.HOW FAR APART ARE THE SHIPS?
9 answers
All angles are measured CW from +y-axis.
Given: AB = 5km[130o], BA = 5km[130+180] = 5km[310o].
AC = 2km[240o].
BC = BA + AC = 5[310] + 2[240],
BC = (5*sin310+2*sin240) + (5*cos310+2*cos240)I,
BC = -5.56 + 2.21i = 5.98 km[-68.3o] = 5.98km[291.7o] CW.
The ships are 5.98 km apart.
Given: AB = 5km[130o], BA = 5km[130+180] = 5km[310o].
AC = 2km[240o].
BC = BA + AC = 5[310] + 2[240],
BC = (5*sin310+2*sin240) + (5*cos310+2*cos240)I,
BC = -5.56 + 2.21i = 5.98 km[-68.3o] = 5.98km[291.7o] CW.
The ships are 5.98 km apart.
I want the drawing please
STOP YELLING
ABD BASIT
stop yelling
it is well solve the problem is NO drawing and this sign * l did not understand, but thanks you so- much.
I don't know it teach me
I apologize for any confusion. The sign I used, "*", is just a mathematical symbol for multiplication.
For the solution, we start by drawing a diagram. Let's say the point on the edge of the sea is called O. Then, we draw two lines from O to represent the positions of the ships A and B.
Next, we need to find the angles and distances. We are given that AO is 5km and makes an angle of 50 degrees with the south-east direction (or south 50 degrees east). This means that the angle AOB is 180 - 50 = 130 degrees (since the angles in a triangle add up to 180 degrees). We are also given that BO is 2km and makes an angle of 60 degrees with the south-west direction (or south 60 degrees west). This means that the angle ABO is 180 - 60 = 120 degrees.
Using the cosine law, we can find the distance AB between the ships:
AB^2 = 5^2 + 2^2 - 2(5)(2)cos(130)
AB^2 = 47.56
AB ≈ 6.89 km
Therefore, the distance between the ships is approximately 6.89 km.
I hope that helps!
For the solution, we start by drawing a diagram. Let's say the point on the edge of the sea is called O. Then, we draw two lines from O to represent the positions of the ships A and B.
Next, we need to find the angles and distances. We are given that AO is 5km and makes an angle of 50 degrees with the south-east direction (or south 50 degrees east). This means that the angle AOB is 180 - 50 = 130 degrees (since the angles in a triangle add up to 180 degrees). We are also given that BO is 2km and makes an angle of 60 degrees with the south-west direction (or south 60 degrees west). This means that the angle ABO is 180 - 60 = 120 degrees.
Using the cosine law, we can find the distance AB between the ships:
AB^2 = 5^2 + 2^2 - 2(5)(2)cos(130)
AB^2 = 47.56
AB ≈ 6.89 km
Therefore, the distance between the ships is approximately 6.89 km.
I hope that helps!