Q65 / 200 · Manipal (MET) 2012 · Mathematics

In a triangle, the angle are in AP and the lengths of the two larger sides are 10 and 9 , respectively, then the length of the third side can be

  • Correct answer
    A
    $5 \pm \sqrt{6}$
  • B
    0.7
  • C
    $15-\sqrt{6}$
  • D
    None of these

Solution

Let the angles be $A=x-d, B=x, C=x+d$, Then, $x-d+x+x+d=180^{\circ}$ $\Rightarrow \quad 3 x=180^{\circ} \Rightarrow x=60^{\circ}$ Therefore, two larger angles are $B$ and $C$. Hence, $b=9$ and $c=10$. Now, $\begin{array}{cc} & \cos B=\frac{c^2+a^2-b^2}{2 a c} \\ \Rightarrow & \frac{1}{2}=\frac{100+a^2-81}{20 a} \\ \Rightarrow & a^2-10 a+19=0 \\ \Rightarrow & a=5 \pm \sqrt{6} \end{array}$

Hence, the correct answer is option: 'A'.

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