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

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

Question

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given $a: b: c = 4: 5: 6$. Let $a = 4k, b = 5k, c = 6k$ for some constant $k > 0$. Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(5k)^2 + (6k)^2 - (4k)^2}{2(5k)(6k)} = \frac{25k^2 + 36k^2 - 16k^2}{60k^2} = \frac{45}{60} = \frac{3}{4}$. $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4k)^2 + (6k)^2 - (5k)^2}{2(4k)(6k)} = \frac{16k^2 + 36k^2 - 25k^2}{48k^2} = \frac{27}{48} = \frac{9}{16}$. $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{(4k)^2 + (5k)^2 - (6k)^2}{2(4k)(5k)} = \frac{16k^2 + 25k^2 - 36k^2}{40k^2} = \frac{5}{40} = \frac{1}{8}$. Now,calculate the expression: $\frac{\cos A + 3 \cos C}{\cos B} = \frac{\frac{3}{4} + 3(\frac{1}{8})}{\frac{9}{16}} = \frac{\frac{3}{4} + \frac{3}{8}}{\frac{9}{16}} = \frac{\frac{6+3}{8}}{\frac{9}{16}} = \frac{\frac{9}{8}}{\frac{9}{16}} = \frac{9}{8} 	imes \frac{16}{9} = 2$.

In $\triangle ABC$,if $a: b: c = 4: 5: 6$,then $\frac{\cos A + 3 \cos C}{\cos B} = $

Similar Questions

In a $\triangle ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then $\triangle ABC$ is

In $\triangle ABC$,if $a^2-c^2=b(b-c)$,$\sqrt{2}a=2b-c$ and $R=\frac{1}{\sqrt{3}}$,then $b=$

In a triangle $ABC$,if $A=30^{\circ}$ and $\frac{b}{(\sqrt{3}+1)^2+2(\sqrt{2}-1)}=\frac{c}{(\sqrt{3}+1)^2-2(\sqrt{2}-1)}$,then $B=$ (in $^{\circ}$)

In a triangle $ABC$,with usual notations,$\cot \left(\frac{A+B}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right) = $

Let $a, b$ and $c$ denote the lengths of sides $BC, CA$ and $AB$ of $\triangle ABC$. In $\triangle ABC$,$\angle BAC = 30^{\circ}$ and $\angle ABC = 60^{\circ}$. Then $a: b: c$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module