In triangle ABC,if b+c : c+a : a+b = 7 : 8 : | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $b+c = 7k$,$c+a = 8k$,and $a+b = 9k$. Adding these equations,we get $2(a+b+c) = 24k$,which implies $a+b+c = 12k$. Subtracting the given equa

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{4}{5}\right)$

Vedclass · Answer

(A) Given $b+c = 7k$,$c+a = 8k$,and $a+b = 9k$. Adding these equations,we get $2(a+b+c) = 24k$,which implies $a+b+c = 12k$. Subtracting the given equations from $a+b+c = 12k$: $a = (a+b+c) - (b+c) = 12k - 7k = 5k$. $b = (a+b+c) - (c+a) = 12k - 8k = 4k$. $c = (a+b+c) - (a+b) = 12k - 9k = 3k$. Since $c Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab}$. $\cos C = \frac{(5k)^2 + (4k)^2 - (3k)^2}{2(5k)(4k)} = \frac{25k^2 + 16k^2 - 9k^2}{40k^2} = \frac{32k^2}{40k^2} = \frac{4}{5}$. Therefore,the smallest angle is $C = \cos^{-1}\left(\frac{4}{5}\right)$.

In $\triangle ABC$,if $b+c : c+a : a+b = 7 : 8 : 9$,then the smallest angle (in radians) of that triangle is

Similar Questions

In any triangle $ABC$,$\frac{\tan \frac{A}{2} - \tan \frac{B}{2}}{\tan \frac{A}{2} + \tan \frac{B}{2}} = $

In any $\triangle ABC$,$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} =$

If $2$ is the length of a side of a triangle with its opposite angle $\frac{\pi}{3}$,then the circumradius of the triangle is ......

In $\triangle ABC$,if $\angle A = 90^{\circ}$,then $2(r+R) = $

If in a triangle $ABC$,$\angle C = 60^{\circ}$,then $\frac{1}{a + c} + \frac{1}{b + c} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module