Aship at position (1, 0) on a nautical chart (with north in the positive y direction) sights a rock at position (7, 5). what angle θ does this vector make with due north? (this is called the bearing of the rock from the ship. round your answer to two decimal places.) answer should be in radians east of north

0.69473828 radians east of north.

Step-by-step explanation:

Interpreting the information given on a Cartesian plane, the position of the ship will be (1,0) and that of the rock will be (7,5). Connecting the point (1,0) and (7,5) with a line and dropping a perpendicular line from point (7,5) to the x-axis to intersect the x-axis at (7,0), will form a right-triangle with base length 6 units and a height of 5 units.

Finding the base length from points (1,0) and (7,0)

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(7-1)²

d= √6²

d=6 units

Finding the height from points (7,5) and (7,0)

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(7-7)² + (0-5)²

d=√-5²

d=√25

d=5 units

So you now have a right triangle with base length 6 units and height 5 units.

To get the bearing of rock from ship position you apply the formula for tangent of an angle

Tan Ф = length of opposite side/length of adjustment side

TanФ=5/6

Tan Ф =0.83333333333

Tan⁻¹ (0.83333333333) =0.69473828 radians east of north.

3: 2

step-by-step explanation:

shorter piece is 2: 3, so just reverse it, and you get 3: 2