Two lighthouses A and B are 25km apart and A is due West of B. A submarine S is on a bearing of 137° from A and on a bearing of 170° from B. Find the distance of S from A and the distance of S from B.

Hence, The distance of S from A is 45.2km and distance of S from B is 33.6km

An illustrative diagram for the scenario is shown in the attachment below

From the diagram,

The distance of S from A is given by /AS/ = b

Consider Δ ASB

/AB/ = 25km, <A = 47° and <B = 100° <S = 33°

To determine /AS/

From Sine rule, we can write that

∴

km

/AS/ ≅ 45.2 km

The distance of S from A is 45.2km

For the distance of S from B

In the triangle, the distance between S and B is given by /SB/ = a

From Sine rule, we can also write that

km

/SB/ ≅ 33.6 km

The distance of S from B is 33.6km

Hence, the distance of S from A is 45.2km and distance of S from B is 33.6km

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SA = 7.97 km

SB = 31.3 km

Step-by-step explanation:

Draw a picture of the triangle formed by the points A, B, and S.

∠SAB = 137°, and ∠ABS = 10°.  Therefore, ∠ASB = 33°.

Using law of sines:

25 / sin 33° = x / sin 10°

x = 7.97

25 / sin 33° = y / sin 137°

y = 31.3

Round as needed.

ok

step-by-step explanation:

the image is given below

since the options are not given so we draw it from the equation

the given equation is

y = (x-1) (x+4)

y = y² + 3x - 4

the above equation is a quadratic equation which will have two roots. if we draw the graph for it, it will look like this