given cos θ= √35/6 and angle θ is in Quadrant I, what is the exact value of sin θ in simplest form? Simplify all radicals if needed.

Here, I've attached a picture of the triangle that accompanies this explanation.

cot = = =

Now view the triangle.

PS. Angle theta will have a coordinate of (-, +) because it is in quadrant II.

sin of the angle would be opposite/hypotenuse, so sin =

cos of the angle would be adjacent/hypotenuse, so cos =

tan of the angle would be opposite/adjacent, so tan =

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

1) is right

step-by-step explanation: