Question

With$11,{\rm{ }}13$,$\sqrt {290 + 143\sqrt 3 }$ as sides
(A) no triangle exists
(B) triangle exists with an angle $\dfrac{{2\pi }}{3}$
(C) triangle exists with an angle $\dfrac{{3\pi }}{4}$
(D) triangle exists with an angle $\dfrac{{5\pi }}{6}$

Answer 1

Hint : First remind the trigonometric formula $\cos \theta = \dfrac{{{a^{2\;}} + {\text{ }}{b^2}-{\text{ }}{c^2}}}{{2ab}}$for the cosine of angle of side with the sides of triangle. Then make the calculations carefully and take the particular solution of the trigonometric equation in the first four quadrants.

Complete step-by-step answer :
With$11,{\rm{ }}13$,$\sqrt {290 + 143\sqrt 3 }$ as sides.
When we are solving this type of question, we need to follow the steps provided in the hint part above.
Let $a = 11,\,\,b = 13$
c = $\sqrt {290 + 143\sqrt 3 }$
Now check

$$a + b = 24 \ge c\\\ b + c = 36.18 \ge a\\\ a + c = 34.18 \ge b$$

Mean triangle is possible. Now we are going to check the angle of triangle by below formula
Let $\theta$ is the angle of triangle
\Rightarrow \cos \theta = $$$$\dfrac{{{a^{2\;}} + {\rm{ }}{b^2}-{\rm{ }}{c^2}}}{{2ab}}

$$= \dfrac{{11 + 13 + \sqrt {290 + 143\sqrt 3 } }}{{2 \times 11 \times 13}}\\\ = \dfrac{{290 - 290 - 143\sqrt 3 }}{{2 \times 11 \times 13}}\\\ = \dfrac{{ - 143}}{{2 \times 143}}\sqrt 3 \\\ \cos \theta = \dfrac{{ - \sqrt 3 }}{2}$$

As $\theta$ if the angle of a triangle so $\theta$ can’t be greater than ${180^0}$
so $\theta$ in 2nd quadrant

$$\Rightarrow \cos \theta = \cos \left( {\pi - \dfrac{\pi }{6}} \right)\\\ \Rightarrow \theta = \left( {\pi - \dfrac{\pi }{6}} \right)\\\ \Rightarrow \theta = \dfrac{{5\pi }}{6}$$

So, the correct answer is “Option D”.

Note : First is take care of the calculations and then make the calculations in the trigonometric formula by substituting values in the formula. Then take care that the solution of the trigonometric equation must be right according to the equation and lies between $0$ and $2\pi$ .