Question

A triangle is formed between boats a and b and a submarine. the length between a and b is 1,425 feet. the angle at point a is 59 degrees and the angle at point b is 47 degrees. ships a and b are 1,425 feet apart and detect a submarine below them. the angle of depression from ship a to the submarine is 59°, and the angle of depression from ship b to the submarine is 47°. how far away is the submarine from the two ships? round to the nearest hundredth of a foot. the distance from ship a to the submarine is about feet. the distance from ship b to the submarine is about feet.

Answer 1

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.

$\frac{a}{sinA}=\frac{b}{sin B} = \frac{c}{sin C}$

so for Boat A and submarine, we use

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

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

When AX is the subject

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

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

AX=1042.245/0.9613

AX=1084.20

The distance between ship B and the submarine

The sine formula we use is

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

Substituting

$\frac{1425}{sin 74}=\frac{BX}{sin 59}$

When BX is the subject

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

= 1425/0.9613 * 0.8572

=1221.51/0.9613

=1270.69

Learn more about the boat and submarines here:

brainly.com/question/17022372

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