Answer:
[tex]AX = 1084.20[/tex]
[tex]BX = 1270.69[/tex]
Step-by-step explanation:
See attachment for complete question
Let the position of the submarine be represented with X
Given
[tex]AB = 1425[/tex]
[tex]\angle A = 59^{\circ}[/tex]
[tex]\angle B = 47^{\circ}[/tex]
First, we calculate angle at X.
[tex]\angle X + \angle A + \angle B = 180[/tex]
[tex]\angle X + 59^{\circ} + 47^{\circ}= 180^{\circ}[/tex]
[tex]\angle X = 180^{\circ} -59^{\circ} - 47^{\circ}[/tex]
[tex]\angle X = 74^{\circ}[/tex]
Solving (a): Distance AX: The distance between ship A and the submarine
To do this, we apply sine formula which states
[tex]\frac{a}{sin\ A} = \frac{b}{sin\ B} = \frac{c}{sin\ C}[/tex]
In this case:
[tex]\frac{AB}{sin\ X} = \frac{AX}{sin\ B}[/tex]
Substitute values for AB, [tex]\angle X[/tex] and [tex]\angle B[/tex]
[tex]\frac{1425}{sin(74^{\circ})} = \frac{AX}{sin(47^{\circ})}[/tex]
Make AX the subject
[tex]AX = \frac{1425}{sin(74^{\circ})} * sin(47^{\circ})[/tex]
[tex]AX = \frac{1425}{0.9613} * 0.7314[/tex]
[tex]AX = \frac{1425 * 0.7314}{0.9613}[/tex]
[tex]AX = \frac{1042.245}{0.9613}[/tex]
[tex]AX = 1084.20[/tex]
Solving (b): Distance BX: The distance between ship B and the submarine
To do this, we apply sine formula which states
In this case:
[tex]\frac{AB}{sin\ X} = \frac{BX}{sin\ A}[/tex]
Substitute values for AB, [tex]\angle X[/tex] and [tex]\angle A[/tex]
[tex]\frac{1425}{sin(74^{\circ})} = \frac{BX}{sin(59^{\circ})}[/tex]
Make BX the subject
[tex]BX = \frac{1425}{sin(74^{\circ})} * sin(59^{\circ})[/tex]
[tex]BX = \frac{1425}{0.9613} * 0.8572[/tex]
[tex]BX = \frac{1425* 0.8572}{0.9613}[/tex]
[tex]BX = \frac{1221.51}{0.9613}[/tex]
[tex]BX = 1270.69[/tex]
Answer:
first answer: 1084.20
second answer: 1270.69
Step-by-step explanation:
correct on edge 2021
