18. The angle of elevation from a viewer to the top of a flagpole is
50°. If the viewer is 20 feet away
and the viewer's eyes are 5 feet
from the ground, how high is the
pole, to the nearest tenth of a foot?

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

just sum

10 + 10 = 20

50-60=10

step-by-step explanation:

3

step-by-step explanation: