Gujarati
Trigonometrical Equations
normal

In a triangle $P Q R, P$ is the largest angle and $\cos P=\frac{1}{3}$. Further the incircle of the triangle touches the sides $P Q, Q R$ and $R P$ at $N, L$ and $M$ respectively, such that the lengths of $P N, Q L$ and $R M$ are consecutive even integers. Then possible length$(s)$ of the side$(s)$ of the triangle is (are)

$(A)$ $16$ $(B)$ $18$ $(C)$ $24$ $(D)$ $22$

A

$(A,D)$

B

$(B,D)$

C

$(B,C)$

D

$(A,C)$

(IIT-2013)

Solution

$\cos P=\frac{(2 n+2)^2+(2 n+4)^2-(2 n+6)^2}{2(2 n+2)(2 n+4)}=\frac{1}{3} $

$\Rightarrow \frac{4 n^2-16}{8(n+1)(n+2)}=\frac{1}{3} $

$=\frac{n^2-4}{2(n+1)(n+2)}=\frac{1}{3} \quad \Rightarrow \quad \frac{n-2}{2(n+1)}=\frac{1}{3} $

$=3 n-6=2 n+2 $

$\Rightarrow n=8 $

$\Rightarrow 2 n+2=18 $

$\Rightarrow 2 n+4=20 $

$\Rightarrow 2 n+6=22$

Standard 11
Mathematics

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