Triangle

[nasi112]

Find the angles of a triangle whose sides have ratios 2:3:4.

 

[Deleted member 4993]

Find the angles of a triangle whose sides have ratios 2:3:4.

Do you know the law of sines of a triangle?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[joshuaa]

let [MATH]a, b, c[/MATH] be the three sides of the triangle and [MATH]A, B, C,[/MATH] the angles opposite to each side respectively

then, by the law of cosines

[MATH]a^2 = b^2 + c^2 - 2bc \cos A[/MATH]
[MATH]A = \cos^{-1} \left(\frac{a^2 - b^2 - c^2}{-2bc}\right)[/MATH]


and you can do the same for the angles [MATH]B[/MATH] and [MATH]C[/MATH]


[MATH]B = \cos^{-1} \left(\frac{b^2 - a^2 - c^2}{-2ac}\right)[/MATH]


[MATH]C = \cos^{-1} \left(\frac{c^2 - a^2 - b^2}{-2ab}\right)[/MATH]


now, all you have to do is to replace [MATH]a, b, c[/MATH] with [MATH]2, 3, 4[/MATH] respectively


[MATH]-\cancel{Jomo} Joshua[/MATH]

 

[Dr.Peterson]

now, all you have to do is to replace [MATH]a, b, c[/MATH] with [MATH]2, 3, 4[/MATH] respectively

Since only ratios, not actual lengths, were given, it's worth pondering why the result obtained for these particular sides will be the same for any triangle with sides proportional to these. It's obvious, but maybe not to newcomers to the subject.

 

[nasi112]

Thank you guys. It is helpful

 

[tkhunny]

Thank you guys. It is helpful

Onw for free, I guess. Perhaps giving you he answer was helpful. Most likely, it was not. Please feel free to follow the forum guidelines and show YOUR work. You can't have nothing to offer. Let's learn some mathematics.