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Ship A and Ship B are 120 km apart when they pick up a distress call from another boat. Ship B estimates that they are 70 km away from the distress call. They also notice that the angle between the line from ship B to ship A and the line from ship A to the distress call is 28°. What are the two possible distances, to the nearest TENTH of a km, from ship A to the boat?

Answer :

CommanderMAM

Answer:

147.5 km and 64.4 km

Step-by-step explanation:

a=120  km

b=70  km

β=28 degrees ( ∘)

 

b^2=(a^2)+(c^2)−2ac*cosβ  

70^2 =(120^2 )+(c^2)−2⋅ 120⋅ c⋅ cos(28∘ )  

 (c^2 ) −211.907c+9500=0  

 

note p, q, and r are replacement variables in the Pythagorean theorem since a, b, and c are already in use

p=1;q=−211.907;r=9500  

D=(q^2 ) −4pr=(211.907^2 )−4⋅1⋅9500=6904.75561996  

D>0  

 

[tex]c_{1,2}[/tex]  =   (−q±  [tex]\sqrt{D}[/tex]   )/2p=(211.91±[tex]\sqrt{6904.76}[/tex])/2

​[tex]c_{1,2}[/tex]  =105.95371114±41.5474295834  

([tex]c_{1}[/tex]−147.501140726)([tex]c_{2}[/tex]−64.4062815596)=0

[tex]c_{1}[/tex]=147.501140726  

[tex]c_{2}[/tex]=64.4062815596  

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