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Sagot :
The distance from ships to the submarine is AX=1084.20
BX=1270.69
Let X be the submarine position.
Given
The length between a and b is AB=1425
and the angle point a is at 59 degrees
the angle point b is at 47 degrees
The calculating angle at X:
∠X+∠A+∠B = 180
∠X+59°+47°=180°
∠X=180°-59°-47°
∠X=74°
Then the distance between boat A and the submarine will be found using sine law
What is sine law?
Sine law is the ratio of each side of a plane triangle to the sine of the opposite angle is the same for all three sides and angles.
[tex]\frac{a}{sinA}=\frac{b}{sin B} = \frac{c}{sin C}[/tex]
so for Boat A and submarine, we use
[tex]\frac{AB}{sin X}=\frac{AX}{sin B}[/tex]
[tex]\frac{1425}{sin (74)} = \frac{AX}{sin (47)}[/tex]
When AX is the subject
AX=[tex]\frac{1425}{sin(74)}*sin(47)[/tex]
=[tex]\frac{1425}{0.9613}*0.7314[/tex]
AX=1042.245/0.9613
AX=1084.20
The distance between ship B and the submarine
The sine formula we use is
[tex]\frac{AB}{sin X}=\frac{BX}{sin A}[/tex]
Substituting
[tex]\frac{1425}{sin 74}=\frac{BX}{sin 59}[/tex]
When BX is the subject
BX=[tex]\frac{1425}{sin 74}*sin(59)[/tex]
= 1425/0.9613 * 0.8572
=1221.51/0.9613
=1270.69
Learn more about the boat and submarines here:
https://brainly.com/question/17022372
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