Question

2 aircraft leave same spot at same time. 650mph W 35° S vs 725mph E 52° S. What distance will there between them in 2hrs?

Answer 1

Answer on Question #58627 – Math – Trigonometry

Question

2 aircraft leave same spot at same time. 650mph W $35{}^{\circ}$ S vs 725mph E $52{}^{\circ}$ S. What distance will there between them in 2hrs?

Solution

$$\angle A = 90{}^{\circ} - 52{}^{\circ} = 38{}^{\circ}$$

The first aircraft flew the following distance:

$$L_1 = L_{red} = 650\,mph \cdot 2\,hrs = 1300\,m.$$

The second aircraft flew the following distance:

$$L_2 = L_{blue} = 725\,mph \cdot 2\,hrs = 1450\,m.$$

The distance $L = L_{green}$ between aircrafts in 2hrs can be found using the law of cosines:

$$\begin{array}{l} L = L_{green} = \sqrt{L_1^2 + L_2^2 - 2L_1L_2\cos\left(L_1; L_2\right)} = \\ = \sqrt{1300^2 + 1450^2 - 2 \cdot 1300 \cdot 1450 \cdot \cos\left(38{}^{\circ} + 90{}^{\circ} + 35{}^{\circ}\right)} \approx 2719.9\,m. \end{array}$$

Answer: 2719.9m.

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