To find third side of a triangle, given angle, then side, then side

[saurabh]

How can we find side BC in given figure?

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[mmm4444bot]

Law of Cosines :cool:

 

[tylercc]

so I am also trying to solve a similar problem. When I do that example I am getting this, What am I doing wrong?

 

[mmm4444bot]

so I am also trying to solve a similar problem. When I do that example I am getting this, What am I doing wrong?
View attachment 8316

Nobody can tell you what you've done wrongly, if you don't show what you did.

Please read the forum guidelines. Thank you. :cool:

 

[tylercc]

well the law of cos says that cos 20 degree is equal to the hypotenuse divided by the opposite. so I made the equation more friendly, a(cos(20))=c. I then put that in to my calculator and I got the answer in the picture. 300(cos(20))=281.91

 

[Deleted member 4993]

How can we find side BC in given figure?

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First use Laws of sines to calculate the angle ACB ← sin(20°)/300 = sin(ACB)/10

Then use ABC + ACB + BAC = 180°

Then use law of cosines ← BC^2 = 300^2 + 10^2 - 2 * 300 * 10 * cos(BAC) ← edited

 

[Deleted member 4993]

well the law of cos says that cos 20 degree is equal to the hypotenuse divided by the opposite. so I made the equation more friendly, a(cos(20))=c. I then put that in to my calculator and I got the answer in the picture. 300(cos(20))=281.91

Law of cosine of a "general" triangle does not say that. Read carefully!!

 

[tylercc]

I was miss understanding the example I was trying to use.

 

[saurabh]

First use Laws of sines to calculate the angle ACB ← sin(20°)/300 = sin(ACB)/10

Then use ABC + ACB + BAC = 180°

Then use law of cosines ← BC^2 = 300^2 + 10^2 - 2 * 300 * 10 * cos(BAC)

Thanks

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[mmm4444bot]

First use Laws of sines to calculate the angle ACB ← sin(20°)/300 = sin(ACB)/10

Then use ABC + ACB + BAC = 180°

Then use law of cosines ← BC^2 = 300^2 + 10^2 - 2 * 300 * 10 * sin(BAC)

You meant cos(BAC)

Or:

300^2 = 10^2 + [BC]^2 - 2*10*[BC]*cos(20°)

 

[Deleted member 4993]

Or:

300^2 = 10^2 + [BC]^2 - 2*10*[BC]*cos(20°)

Yes indeed - I edited my post. Thanks...

 

[tkhunny]

Let us also remember that Angle-Side-Side does not necessarily a triangle make. Just because someone drew it, doesn't mean it exists. Always check!

 

[stapel]

Let us also remember that Angle-Side-Side does not necessarily a triangle make....

Which is why some instructors highlight the issue by referring to this sort of triangle as \(\displaystyle \, \mbox{A}\mbox{SS.}\) :lol: