Exams Joint Entrance Examination Mathematics Properties of Triangle Question 22 Question 22 of 26 Medium MCQ
In a triangle ABC, if \[\cos \mathrm{A}+2 \cos \mathrm{B}+\cos C=2\] and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then \[\mathrm{\cos A-\cos C}\] is equal to
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\[
cos A + 2 cos B + cos C = 2 cos A + cos C = 2 ( 1 − cos B ) 2 cos A + C 2 cos ( A − C 2 ) = 2 × 2 sin 2 B 2 cos A − C 2 = 2 sin B 2 2 cos B 2 cos A − C 2 = 4 sin B 2 cos B 2 2 sin ( A + C 2 ) cos ( A − C 2 ) = 2 sin B sin A + sin C = 2 sin B a + c = 2 b ( ∵ a = 3 , c = 7 ) ⇒ b = 5 \begin{aligned}
& \cos A+2 \cos B+\cos C=2 \\\\
& \cos A+\cos C=2(1-\cos B) \\\\
& 2 \cos \frac{A+C}{2} \cos \left(\frac{A-C}{2}\right)=2 \times 2 \sin ^2 \frac{B}{2} \\\\
& \cos \frac{A-C}{2}=2 \sin \frac{B}{2} \\\\
& 2 \cos \frac{B}{2} \cos \frac{A-C}{2}=4 \sin \frac{B}{2} \cos \frac{B}{2} \\\\
& 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)=2 \sin B \\\\
& \sin A+\sin C=2 \sin B \\\\
& a+c=2 b \quad(\because a=3, c=7) \\\\
& \Rightarrow b=5 \quad
\end{aligned} cos A + 2 cos B + cos C = 2 cos A + cos C = 2 ( 1 − cos B ) 2 cos 2 A + C cos ( 2 A − C ) = 2 × 2 sin 2 2 B cos 2 A − C = 2 sin 2 B 2 cos 2 B cos 2 A − C = 4 sin 2 B cos 2 B 2 sin ( 2 A + C ) cos ( 2 A − C ) = 2 sin B sin A + sin C = 2 sin B a + c = 2 b ( ∵ a = 3 , c = 7 ) ⇒ b = 5
\]
\[
cos A − cos C = b 2 + c 2 − a 2 2 b c − a 2 + b 2 − c 2 2 a b = 25 + 49 − 9 70 − 9 + 25 − 49 30 = 65 70 + 1 2 = 20 14 = 10 7 \begin{aligned}
\cos A & -\cos C=\frac{b^2+c^2-a^2}{2 b c}-\frac{a^2+b^2-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} cos A − cos C = 2 b c b 2 + c 2 − a 2 − 2 ab a 2 + b 2 − c 2 = 70 25 + 49 − 9 − 30 9 + 25 − 49 = 70 65 + 2 1 = 14 20 = 7 10
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