Show that only one n, n+2 and n+4 are divisible by 3
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Answer:
steps below
Step-by-step explanation:
3.2.1 AD = DB* sin 2 = DB * sin θ .. DE // AB ∠2= θ ... (1)
By laws of sines: DB / sin ∠5 = x / sin ∠4
∠4 = θ-α ∠5 = 180°-<u>∠1</u>-∠4 = 180°-<u>∠3</u>-∠4 = 180°-(90°-θ)-(θ-α)) = 90°+α
DB = (x*sin ∠5)/sin (θ-α)
= (x* sin (90°+α)) / sin (θ-α)
AD = DB*sinθ
= (x* sin (90°+α))*sinθ / sin (θ-α)
= x* (sin90°cosα+cos90°sinα)*sinθ / sin (θ-α) .... sin90°=1, cos90°=0
= x* cosα* sinθ / sin (θ-α)
3.2.2 Please apply Laws of sines to calculate the length
