Question

From a lighthouse L an aircraft carrier A is 15km away on a bearing of 112 degrees and a submarine S is 26 km away on a bearing of 200 degrees. Find
a) the distance between A and S
b) the bearing of A from S

Answer 1

Answer on Question #62826 – Math – Trigonometry

Question

From a lighthouse L an aircraft carrier A is 15km away on a bearing of 112 degrees and a submarine S is 26 km away on a bearing of 200 degrees.

Find

a) the distance between A and S

b) the bearing of A from S

Solution

a) Compute $\angle ALS = 200{}^{\circ} - 112{}^{\circ} = 88{}^{\circ}$.

Using the law of cosines one finds side AS:

$$\begin{array}{l} c^{2} = a^{2} + b^{2} - 2ab \cdot \cos(88{}^{\circ}); \\ c^{2} = 15^{2} + 26^{2} - 2 \cdot 15 \cdot 26 \cdot \cos(88{}^{\circ}); \\ c^{2} = 225 + 676 - 780 \cdot 0.0348995; \\ c^{2} = 873.778; \\ c = 29.5597; \end{array}$$

So the distance between A and S is 29.5597 km.

b) Now find the angle $\angle LAS$ by the law of sines:

$$\begin{array}{l} \frac{LS}{\sin \angle LAS} = \frac{AS}{\sin \angle ALS}; \\ \frac{\sin(\angle LAS)}{26} = \frac{\sin(88{}^{\circ})}{29.5597}; \\ \sin(\angle LAS) = 26 \cdot \frac{\sin(88{}^{\circ})}{29.5597}; \\ \angle LAS = \asin\left(26 \cdot \frac{\sin(88{}^{\circ})}{29.5597}\right), \end{array}$$

where asin is the inverse of the sine function;

$$\angle LAS = \asin(0.8790400952);$$ $$\angle LAS = 61.53{}^{\circ}.$$

The bearing of A from S is angle $\angle ASB$:

$$\angle ASB = \angle ASL + \angle LSB;$$

Now find the angle $\angle ASL$:

$$\angle ASL = 180{}^{\circ} - \angle ALS - \angle LAS = 180{}^{\circ} - 88{}^{\circ} - 61.53{}^{\circ} = 30.47{}^{\circ}.$$

As lines CL and SB are parallel, we know that alternate interior angles are equal, so

$$\angle LSB = \angle SLC = 200{}^{\circ} - 180{}^{\circ} = 20{}^{\circ}.$$

Then

$$\angle ASB = \angle ASL + \angle LSB = 30.47{}^{\circ} + 20{}^{\circ} = 50.47{}^{\circ}.$$

Answer: the distance between A and S is $29.5597\,\mathrm{km}$, the bearing of A from S is $50.47{}^{\circ}$.

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