Question

two weather stations are aware of a thunderstorm located at point c. the weather stations a and b are 24 miles apart. how far is weather station a from the storm? 44.6 miles 19.7 miles 42.2 miles 31.2 miles

Explanation:

Step1: Find angle at C

First, calculate the internal angles at A and B:
Angle at A: $180^\circ - 17^\circ = 163^\circ$
Angle at B: $180^\circ - 48^\circ = 132^\circ$
Sum of angles in a triangle is $180^\circ$, so:
$\angle C = 180^\circ - 163^\circ - 132^\circ + 180^\circ$ (correcting for external angles: actual internal angles sum to $180^\circ$, so $\angle C = 180^\circ - (180^\circ-17^\circ) - (180^\circ-48^\circ) = 17^\circ + 48^\circ - 180^\circ + 180^\circ = 65^\circ$)
$\angle C = 180^\circ - (180^\circ-17^\circ) - (180^\circ-48^\circ) = 17^\circ + 48^\circ - 180^\circ + 180^\circ = 65^\circ$

Step2: Apply Law of Sines

Let $AC = x$ (distance we need to find), $AB = 24$ miles.
Law of Sines: $\frac{x}{\sin(\angle B_{\text{internal}})} = \frac{AB}{\sin(\angle C)}$
Internal angle at B: $180^\circ - 48^\circ = 132^\circ$
$\frac{x}{\sin(132^\circ)} = \frac{24}{\sin(65^\circ)}$

Step3: Solve for x

$x = \frac{24 \times \sin(132^\circ)}{\sin(65^\circ)}$
$\sin(132^\circ) = \sin(48^\circ) \approx 0.7431$, $\sin(65^\circ) \approx 0.9063$
$x = \frac{24 \times 0.7431}{0.9063} \approx \frac{17.8344}{0.9063} \approx 19.7$

Answer:

19.7 miles