Don't try these without a diagram.
Draw a straight line (the ocean) and mark two points A and B (the two ships)
put a point above that line and to the left of points A and B, name that point P for the plane.
Draw a perpendicular from P to the line containing A and B and call that point Q
You should now see two right angled triangles, PQA and PQB, with the right angle at Q
From the description angle PAQ = 8.33º and angle PBQ = 75.67º and PQ = 2
in the large triangle,
tan 8.33 = 2/QB
QB = 2/tan 8.33
in the smaller triangle
tan 75.67 = 2/QA
QA = 2/tan 75.67
so AB = QB - QA
= 2/tan 8.33 - 2/tan 75.67
Calculator time !
I have no idea how to do this problem
From an airplane flying 2 kilometers aboce the ocean, a piolet sees two ships directly to the east. The angles of depression to the ships are 8 degrees 20 minutes and 75 degrees 40 minutes how far away are the ships?
I know what the anlge of depression is the only thing is that my teach told me the answer was 13.03 and I got like 3 something...
I don\'t think I know how to properly do this problem
4 answers
Let A be the angle of depression of a ship and let H be the altitude of the airplane.
The distance of each ship from the vertical below the airplane, measured at sea level, is
D = H/tan A
That distance is 13.65 miles for the ship with 8.333 degree depression angle, and 0.51 miles for the other ship. The distance BETWEEN the ships (which is not what you asked for) is 13.14 miles
The distance to the ships from the airplane is
H/(sin A1) = 13.80 km
H/(sin A2) = 2.06 km
The distance of each ship from the vertical below the airplane, measured at sea level, is
D = H/tan A
That distance is 13.65 miles for the ship with 8.333 degree depression angle, and 0.51 miles for the other ship. The distance BETWEEN the ships (which is not what you asked for) is 13.14 miles
The distance to the ships from the airplane is
H/(sin A1) = 13.80 km
H/(sin A2) = 2.06 km
You should now see two right angled triangles, PQA and PQB, with the right angle at Q
From the description angle PAQ = 8.33º and angle PBQ = 75.67º and PQ = 2
were did you get the angles from
I thought you could do
90 - 75 degrees 40 minutes to get the angle QPA but I got 14
From the description angle PAQ = 8.33º and angle PBQ = 75.67º and PQ = 2
were did you get the angles from
I thought you could do
90 - 75 degrees 40 minutes to get the angle QPA but I got 14
The angle PAQ is the angle of elevation viewed from the ship, which is the same as the angle of depression viewed from the airplane (8°20'). The same for ship B.
This is because the angles of elevation and depression are alternate angles between two parallel lines.
I agree with the two calculations above using H/tan(θ) for the horizontal distance, and the answer of 13.143 km.
This is because the angles of elevation and depression are alternate angles between two parallel lines.
I agree with the two calculations above using H/tan(θ) for the horizontal distance, and the answer of 13.143 km.
1 answer