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Answer :
To find the distance between locations [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we can use the information about the angles and one known side in the triangle formed by locations [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex].
1. Identify the Known Values:
- The distance from [tex]\(C\)[/tex] to [tex]\(B\)[/tex] is 15 miles.
- [tex]\(\angle B = 105^\circ\)[/tex].
- [tex]\(\angle C = 20^\circ\)[/tex].
2. Calculate [tex]\(\angle A\)[/tex]:
Since the sum of angles in a triangle is always [tex]\(180^\circ\)[/tex], we can find [tex]\(\angle A\)[/tex] as follows:
[tex]\[
\angle A = 180^\circ - \angle B - \angle C = 180^\circ - 105^\circ - 20^\circ = 55^\circ
\][/tex]
3. Use the Law of Sines:
The Law of Sines states that in any triangle:
[tex]\[
\frac{a}{\sin A} = \frac{c}{\sin C}
\][/tex]
where [tex]\(a\)[/tex] is the side opposite [tex]\(\angle A\)[/tex] (the distance between [tex]\(A\)[/tex] and [tex]\(B\)[/tex]) and [tex]\(c\)[/tex] is the side opposite [tex]\(\angle C\)[/tex] (the distance [tex]\(CB\)[/tex]).
4. Calculate the Distance [tex]\(AB\)[/tex]:
Rearrange the Law of Sines to solve for [tex]\(a\)[/tex]:
[tex]\[
a = \frac{c \cdot \sin A}{\sin C}
\][/tex]
Substituting the known values:
[tex]\[
a = \frac{15 \cdot \sin(55^\circ)}{\sin(20^\circ)}
\][/tex]
Evaluating the sines and performing the division, we get:
[tex]\[
a \approx 35.9 \text{ miles}
\][/tex]
Hence, the distance between locations [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is approximately [tex]\(35.9\)[/tex] miles, to the nearest tenth of a mile.
1. Identify the Known Values:
- The distance from [tex]\(C\)[/tex] to [tex]\(B\)[/tex] is 15 miles.
- [tex]\(\angle B = 105^\circ\)[/tex].
- [tex]\(\angle C = 20^\circ\)[/tex].
2. Calculate [tex]\(\angle A\)[/tex]:
Since the sum of angles in a triangle is always [tex]\(180^\circ\)[/tex], we can find [tex]\(\angle A\)[/tex] as follows:
[tex]\[
\angle A = 180^\circ - \angle B - \angle C = 180^\circ - 105^\circ - 20^\circ = 55^\circ
\][/tex]
3. Use the Law of Sines:
The Law of Sines states that in any triangle:
[tex]\[
\frac{a}{\sin A} = \frac{c}{\sin C}
\][/tex]
where [tex]\(a\)[/tex] is the side opposite [tex]\(\angle A\)[/tex] (the distance between [tex]\(A\)[/tex] and [tex]\(B\)[/tex]) and [tex]\(c\)[/tex] is the side opposite [tex]\(\angle C\)[/tex] (the distance [tex]\(CB\)[/tex]).
4. Calculate the Distance [tex]\(AB\)[/tex]:
Rearrange the Law of Sines to solve for [tex]\(a\)[/tex]:
[tex]\[
a = \frac{c \cdot \sin A}{\sin C}
\][/tex]
Substituting the known values:
[tex]\[
a = \frac{15 \cdot \sin(55^\circ)}{\sin(20^\circ)}
\][/tex]
Evaluating the sines and performing the division, we get:
[tex]\[
a \approx 35.9 \text{ miles}
\][/tex]
Hence, the distance between locations [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is approximately [tex]\(35.9\)[/tex] miles, to the nearest tenth of a mile.
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