Astraight river flows east at a speed of 8 mi/h. a boater starts at the south shore of the river and wants to arrive at a point on the north shore of the river directly opposite the starting point. the motorboat has a speed of 16 mi/h relative to the water. in what direction should the boat be headed?

specifically 30º west of the north

Explanation:

This exercise can be solved using the composition of vectors, let's call the speed of the river that goes east (v₁ = 8 mi / h) to the speed of the boat (v₂ = 16 mi / h) and the desired speed from south to north (v₃ =?).  Look at the attachment to be clear about the vectors

Let's use the Pythagorean triangle

    v₃² = v₁² + v₂²

Let's  calculate v₂

    v₂² = v₃² - v₁²

    v₂ = √ (16² - 8²)

    v₂ = 13.86 mi / h

To find the angle we use trigonometry

    sin θ = v1 / v3

    θ = sin⁻¹ (v1 / v3)

    θ = sin⁻¹ (8/16)

    θ = 30º

The boat must be pointed at 30º with respect to the south-north direction, specifically 30º west of the north

photocells convert light energy to electricity

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement