Question

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Answer 1

Answer:

let 30 ft. be the base.

Area of parallelogram = base * height

300 sq. ft. = 30 ft. * h

h = 300/30 ft. => h = 10 ft.

draw BL perpendicular to AD.

Now, in ΔABL,

∠BLA = 90°

sin∠A = opp./hyp.

          = 10/20=1/2

sin∠A = sin 60°

∴ ∠A = 60° and ∠A=∠C (Opp. angles in a parallelogram are equal)

∠A +  ∠D = 180°

∴∠D = 120° and ∠D=∠B (Adjacent angles in a parallelogram are supplementary)

∴ ∠A = 60°, ∠B = 120°, ∠C = 60°, ∠D = 120°

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