A ladder 10 m long, leans against a vertical wall at an angle of 70° to the ground. if the ladder slips down the wall 4m, find, correct to 2 significant figure (a) the new angle which the ladder makes with the ground

(b) the distance the ladder slipped back on the ground from it's original position

(a) The new angle the ladder makes with the ground = 32.66°

(b) The distance the ladder slipped back on the ground from it's original position = 5 m

Step-by-step explanation:

The given parameters are;

The length of the ladder = 10 m

Angle of inclination of the ladder = 70°

Length to which the ladder slips down the wall = 4 m

(a) The height of the wall adjacent to the ladder = 10×sin(70°) ≈ 9.4 m

The new height of the wall adjacent to the ladder after the ladder slips down 4 m = 9.4 - 4 = 5.4 m

Therefore, the new angle of inclination of the ladder = sin⁻¹(5.4/10) = 32.66°

The new angle the ladder makes with the ground = 32.66°

(b) The distance the ladder slipped back on the ground from its original position is found as follows

The initial distance of the base of the ladder from the wall  = 10 × cos(70°) = 3.42 m

The distance of the base of the ladder from the wall after slipping 4 m down the wall = 10 × cos(32.66°)  ≈ 8.42 m.

Therefore, the amount by which the ladder slip back on the ground from its original position = 8.42 - 3.42 = 5 m.

The distance the ladder slipped back on the ground from it's original position = 5 m.

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: