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sergeinik [125]
3 years ago
10

Ships a and b or 1425 feet apart and detect a submarine below them

Mathematics
1 answer:
Verizon [17]3 years ago
7 0

Answer:

AX = 1084.20

BX = 1270.69

Step-by-step explanation:

See attachment for complete question

Let the position of the submarine be represented with X

Given

AB = 1425

\angle A = 59^{\circ}

\angle B = 47^{\circ}

First, we calculate angle at X.

\angle X + \angle A + \angle B = 180

\angle X + 59^{\circ} + 47^{\circ}= 180^{\circ}

\angle X = 180^{\circ} -59^{\circ} - 47^{\circ}

\angle X = 74^{\circ}

Solving (a): Distance AX: The distance between ship A and the submarine

To do this, we apply sine formula which states

\frac{a}{sin\ A} = \frac{b}{sin\ B} = \frac{c}{sin\ C}

In this case:

\frac{AB}{sin\ X} = \frac{AX}{sin\ B}

Substitute values for AB, \angle X and \angle B

\frac{1425}{sin(74^{\circ})} = \frac{AX}{sin(47^{\circ})}

Make AX the subject

AX = \frac{1425}{sin(74^{\circ})} * sin(47^{\circ})

AX = \frac{1425}{0.9613} * 0.7314

AX = \frac{1425 * 0.7314}{0.9613}

AX = \frac{1042.245}{0.9613}

AX = 1084.20

Solving (b): Distance BX: The distance between ship B and the submarine

To do this, we apply sine formula which states

In this case:

\frac{AB}{sin\ X} = \frac{BX}{sin\ A}

Substitute values for AB, \angle X and \angle A

\frac{1425}{sin(74^{\circ})} = \frac{BX}{sin(59^{\circ})}

Make BX the subject

BX = \frac{1425}{sin(74^{\circ})} * sin(59^{\circ})

BX = \frac{1425}{0.9613} * 0.8572

BX = \frac{1425* 0.8572}{0.9613}

BX = \frac{1221.51}{0.9613}

BX = 1270.69

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