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  • In a triangle ABC, if cos A + 2 cos B + cos C = 2 and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then cos A – cos C is equal to<div class='qn

In a triangle ABC, if cos A + 2 cos B + cos C = 2 and the lengths of the sides opposite to the angles A and C
are 3 and 7 respectively, then cos A – cos C is equal to

Option: 1

\frac{5}{7}


Option: 2

\frac{9}{7}


Option: 3

\frac{3}{7}


Option: 4

\frac{10}{7}


Answers (1)

best_answer

\begin{aligned} & \cos \mathrm{A}+\cos \mathrm{C}=2(1-\cos \mathrm{B}) \\ & 2 \cos \frac{\mathrm{A}+\mathrm{C}}{2} \cos \frac{\mathrm{A}-\mathrm{C}}{2}=4 \sin ^2 \mathrm{~B} / 2 \\ & \text { as } \cos \left(\frac{\mathrm{A}+\mathrm{C}}{2}\right)=\sin \frac{\mathrm{B}}{2} \\ & \text { so } \cos \frac{\mathrm{A}-\mathrm{C}}{2}=2 \sin \frac{\mathrm{B}}{2} \\ & 2 \cos \mathrm{B} / 2 \cos \frac{\mathrm{A}-\mathrm{C}}{2}=4 \sin \mathrm{B} / 2 \cos \mathrm{B} / 2 \\ & 2 \sin \left(\frac{\mathrm{A}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{C}}{2}\right)=4 \sin \mathrm{B} / 2 \cos \mathrm{B} / 2 \\ & \sin \mathrm{A}+\sin \mathrm{C}=2 \sin \mathrm{B} \\ & \mathrm{a}+\mathrm{c}=2 \mathrm{~b} \Rightarrow \mathrm{a}=3, \mathrm{c}=7, \mathrm{~b}=5 \end{aligned}

\begin{aligned} & \cos \mathrm{A}-\cos \mathrm{C}=\frac{\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2}{2 \mathrm{bc}}-\frac{\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2}{2 \mathrm{ab}} \\ & =\frac{25+49-9}{70}-\frac{9+25-49}{30} \\ & \frac{65}{70}+\frac{1}{2}=\frac{20}{14}=\frac{10}{7} \end{aligned}

 

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