Bearing

A plane flew 50 km on a bearing 63°20' and the flew on a bearing 153°20' for 140km. Find the distance between the starting point and the ending point.

Correct answer:

s = 148.6607 km

Step-by-step explanation:

A = 63 \frac{2 0}{6 0} = 63 + \frac{2 0}{6 0} = \frac{6 3 \cdot 6 0 + 2 0}{6 0} = \frac{3 8 0 0}{6 0} = \frac{1 9 0}{3} ≐ 63.3333^{\circ} B = 153 \frac{2 0}{6 0} = 153 + \frac{2 0}{6 0} = \frac{1 5 3 \cdot 6 0 + 2 0}{6 0} = \frac{9 2 0 0}{6 0} = \frac{4 6 0}{3} ≐ 153.3333^{\circ} a = 50 km b = 140 km x_{1} = a \cdot cos A = a \cdot cos 63.333333333333 ° = 50 \cdot cos 63.333333333333 ° = 50 \cdot 0.448799 = 22.43996 km y_{1} = a \cdot sin A = a \cdot sin 63.333333333333 ° = 50 \cdot sin 63.333333333333 ° = 50 \cdot 0.893633 = 44.68163 km x_{2} = b \cdot cos B = b \cdot cos 153.33333333333 ° = 140 \cdot cos 153.33333333333 ° = 140 \cdot (- 0.893633) = - 125.10857 = (- 229422 π / 5761) km y_{2} = b \cdot sin B = b \cdot sin 153.33333333333 ° = 140 \cdot sin 153.33333333333 ° = 140 \cdot 0.448799 = 62.83189 km x = x_{1} + x_{2} = 22.44 + (- 125.1086) ≐ - 102.6686 km y = y_{1} + y_{2} = 44.6816 + 62.8319 ≐ 107.5135 km s = x^{2} + y^{2} = (- 102.6686)^{2} + 107.513 5^{2} = 10221 = 148.6607 km

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