Question

If the sides of the triangles are $p,q,\sqrt{\left( {{p}^{2}}+{{q}^{2}}+pq \right)}$ , then the greatest angle is
$A)\dfrac{\pi }{2}$
$B)\dfrac{5\pi }{2}$
$C)\dfrac{2\pi }{3}$
$D)\dfrac{7\pi }{4}$

Answer 1

To solve the question we need to have the knowledge of trigonometric function$\cos \theta$ and the formula to find an angle in the form of $\cos$ inverse. So the first step will be to write the formula of cos theta in the algebraic terms. The second step will be to find the angle by taking the inverse of the algebraic term.

Complete step by step answer:
The question ask us to find the largest angle when the sides of the triangle is given as $p,q,\sqrt{\left( {{p}^{2}}+{{q}^{2}}+pq \right)}$. The first step will be to find the angle of $\cos \theta$ using the formula. For this we will be finding the ratio of the difference of the square of the third side of the triangle from the square of the sum of the left two sides to the twice of the product of the other sides. On writing it mathematically we get:
$\Rightarrow \cos \theta =\dfrac{{{p}^{2}}+{{q}^{2}}-{{\sqrt{{{p}^{2}}+{{q}^{2}}+pq}}^{2}}}{2pq}$
On solving the above equation we get:
$\Rightarrow \cos \theta =\dfrac{{{p}^{2}}+{{q}^{2}}-\left( {{p}^{2}}+{{q}^{2}}+pq \right)}{2pq}$
As we know that the multiplication of positive and negative is always negative, applying the same in the above expression we get:
$\Rightarrow \cos \theta =\dfrac{{{p}^{2}}+{{q}^{2}}-{{p}^{2}}-{{q}^{2}}-pq}{2pq}$
From the above expression we see that the terms in the numerator get cancelled leaving $-pq$ in the numerator. So the solution thus become:
$\Rightarrow \cos \theta =\dfrac{-pq}{2pq}$
$\Rightarrow \cos \theta =\dfrac{-1}{2}$
To find the angle $\theta$ , we will find the “cos” inverse of the given function. On doing so we get:
$\Rightarrow {{\cos }^{-1}}\left( \cos \theta \right)={{\cos }^{-1}}\left( \dfrac{-1}{2} \right)$
On simplifying we get:
$\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{-1}{2} \right)$
Since the domain is negative, this means the angle is between $\dfrac{\pi }{2}$ to $\dfrac{3\pi }{2}$. On calculating we get:
$\Rightarrow \theta =\dfrac{2\pi }{3}$
$\therefore$ If the sides of the triangles are $p,q,\sqrt{\left( {{p}^{2}}+{{q}^{2}}+pq \right)}$ , then the greatest angle is $\dfrac{2\pi }{3}$ .

So, the correct answer is “Option C”.

Note: To solve these types of questions we should remember the formulas for trigonometric functions and their inverse function. We should know that the trigonometric functions $\sin$ and $\text{cosec}$ are positive of the angles in ${{1}^{st}}$ and ${{2}^{nd}}$ quadrant. Similarly that the trigonometric functions $\tan$ and $\text{cot}$ are positive of the angles in ${{1}^{st}}$ and ${{3}^{rd}}$ quadrant and the trigonometric functions $\cos$ and $\sec$ are positive of the angles in ${{1}^{st}}$ and ${{4}^{th}}$ quadrant.