In triangle ABC,if a: b: c = 4: 5: 6,then cos | vedclass

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

Question

(A) Given the sides $a: b: c = 4: 5: 6$,let $a = 4k$,$b = 5k$,and $c = 6k$ for some constant $k > 0$. Using the Law of Cosines: $\cos A = \frac{b^2 +

Vedclass · Accepted Answer

$12: 9: 2$

Vedclass · Answer

(A) Given the sides $a: b: c = 4: 5: 6$,let $a = 4k$,$b = 5k$,and $c = 6k$ for some constant $k > 0$. Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(5k)^2 + (6k)^2 - (4k)^2}{2(5k)(6k)} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = \frac{3}{4}$. $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4k)^2 + (6k)^2 - (5k)^2}{2(4k)(6k)} = \frac{16 + 36 - 25}{48} = \frac{27}{48} = \frac{9}{16}$. $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{(4k)^2 + (5k)^2 - (6k)^2}{2(4k)(5k)} = \frac{16 + 25 - 36}{40} = \frac{5}{40} = \frac{1}{8}$. Now,find the ratio $\cos A : \cos B : \cos C = \frac{3}{4} : \frac{9}{16} : \frac{1}{8}$. Multiply by the least common multiple of the denominators,which is $16$: $\cos A : \cos B : \cos C = (\frac{3}{4} 	imes 16) : (\frac{9}{16} 	imes 16) : (\frac{1}{8} 	imes 16) = 12 : 9 : 2$.

In $\triangle ABC$,if $a: b: c = 4: 5: 6$,then $\cos A: \cos B: \cos C =$

Similar Questions

In a $\triangle ABC$,$A = 30^{\circ} + C$ and $R = (\sqrt{3} + 1)r$,where $r$ is the inradius and $R$ is the circumradius. Then:

In $\triangle ABC$,if $(a+c)^2 = b^2 + 3ca$,then $\frac{a+c}{2R} =$

The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one,then the sides of the triangle (in units) are

If $a = 2, b = 3, c = 5$ in $\Delta ABC$,then $C = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module