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It is 4.7 km from Lighthouse A to Port B. The bearing of the port from the lighthouse is N73°E. A ship has sailed due west from the port and its bearing from the
lighthouse is N31°E. How far has the ship sailed from the port? Round your answer to the nearest 0.1 km
O A. 3.5 km
O B. 3.1 km
O C. 3.7 km​

Respuesta :

Answer:

correct option is  C. 3.7 km

Step-by-step explanation:

given data

A to Port B = 4.7 km

lighthouse = N73°E

lighthouse = N31°E

solution

we get here first  [tex]\angle[/tex] B and  

here

[tex]\angle[/tex] A = 90 - 73 = 17°

[tex]\angle[/tex] B =  73 - 31 = 42°

and

sum of all angle 180° so

[tex]\angle[/tex] A +  

17° + 42° +  [tex]\angle[/tex] C = 180°  

solve it we get

[tex]\angle[/tex] C = 121°

Now we use here sin law that is

[tex]\frac{b}{sinB} = \frac{c}{sinC}[/tex]   ........................2

put here value and we get

[tex]\frac{b}{sin42} = \frac{4.7}{sin121}[/tex]  

solve it we get

b = 3,7 km

so correct option is  C. 3.7 km

Ver imagen DeniceSandidge
aksnkj

The ship has sailed about 3.7 kilometers. Option C is correct.

Given information:

The distance between Lighthouse A and port B is AB= z = 4.7 km.

The bearing of the port from the lighthouse is N73°E.

A ship has sailed due west from the port and its bearing from the  lighthouse is N31°E.

See the attached figure with triangle ABC.

From triangle ABC,

[tex]\angle A=90-73=17^{\circ}\\\angle B=73-31=42^{\circ}[/tex]

Use the angle sum property to get the value of angle C as,

[tex]\angle C=180-17-42\\=121^{\circ}[/tex]

Let the length of sides of triangle be x, y, and z as shown in the image.

Use sine law to get the value of distance the ship has sailed,

[tex]\dfrac{y}{sinB}=\dfrac{z}{sinC}\\\dfrac{y}{sin42}=\dfrac{4.6}{sin121}\\y=3.6689\\y\approx 3.7[/tex]

Therefore, the ship has sailed about 3.7 kilometers.

For more details, refer to the link:

https://brainly.com/question/22225513

Ver imagen aksnkj

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