In any triangle ABC , the sides are 6 cm , 10 cm and 14 cm . Then the triangle is obtuse angled with the obtuse angle equal to

• A

  $150^{\circ}$
• B

  $135^{\circ}$
• Correct answer
  C

  $120^{\circ}$
• D

  $105^{\circ}$

Solution

$\begin{aligned} \cos \mathrm{C} & =\frac{a^2+b^2-c^2}{2 a b}=\frac{36+100-196}{2(6)(10)} \\ & =\frac{-60}{120}=\frac{-1}{2} \\ \cos \mathrm{C} & =\cos 120^{\circ} \\ \mathrm{C} & =120^{\circ} .\end{aligned}$

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