The wind is blowing N 35.0 degrees W at 1.60 * 10^2 mph. A plane has an engine speed of 3.20 * 10^2 mph. Where should the pilot point the plane in order to fly straight W?

24.2° SouthWest

Step-by-step explanation:

First step: DRAW A PICTURE of the vectors from head to tail (see image)

I created a perpendicular from the resultant vector to the vertex of the given vectors so I could use Pythagorean Theorem to find the length of the perpendicular. Then I used that value to find the angle of the plane.

Perpendicular (x):

 cos 35° = adjacent/hypotenuse

  cos 35° = x/160

→ x = 160 cos 35°

Angle (θ):

sin θ = opposite/hypotenuse

sin θ = x/320

sin θ = 160 cos 35°/320

    θ = arcsin (160 cos 35°/320)

    θ = 24.2°

Direction is down (south) and left (west)

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

step-by-step explanation:

hihi so drawing these 2 triangles is fairly simple. all that's left is finding the height of each one. luckily, we have a side length and an angle, so there is a trig function we can use to solve for the height. the easiest one to use would be tangent, since we have 2/3 things needed to find and the third needed is what we are solving for. tan(73°) * 30 ~= 98.13 and tan(73°) * 35 ~= 114.45