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Home » We know $ an 60^\circ = \sqrt3 $, so: - Blask

We know $ an 60^\circ = \sqrt3 $, so: - Blask

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 35% of 120: 120 × 0.35 = 42 → correct.

📌 − 42 = 78 → correct.

📌 1/4 of 78 = 78 / 4 = 19.5 → still.

📌 But in such problems, we assume exact values, so perhaps the fraction is symbolic.

📌 But final answer: number present = 78 − 19.5 = 58.5 → invalid.

📌 Wait — mistake: the problem says 1/4 are absent — so 1/4 away, so present = 3/4 of girls.

📌 So 3/4 × 78 = (3 × 78) / 4 = 234 / 4 = 58.5 → still fractional.

📌 But 78 × 3 / 4 = 234 / 4 = 117 / 2 = 58.5 → not integer.

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Break first leg (4.2 km NE):
x₁ = 4.2 × sin(45°) ≈ 4.2 × 0.7071 ≈ <<4.2*0.7071=2.97042>>2.97 km (ea...
y₁ = 4.2 × cos(45°) ≈ 4.2 × 0.7071 ≈ <<4.2*0.7071=2.97042>>2.97 km (no...
Second leg (5.6 km SE):
x₂ = 5.6 × sin(45°) ≈ 5.6 × 0.7071 ≈ <<5.6*0.7071=3.96396>>3.96 km (ea...
y₂ = −5.6 × cos(45°) ≈ −5.6 × 0.7071 ≈ <<−5.6*0.7071=−3.96396>>−3.96 k...
Total displacement:
x_total = 2.97 + 3.96 ≈ <<2.97+3.96=6.93>>6.93 km east,
y_total = 2.97 − 3.96 ≈ <<2.97-3.96=−0.99>>−0.99 km ≈ 0.99 km south.

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