# A triangle having vertices as $P\left( \cos \alpha ,\sin \alpha \right),Q\left( \cos \beta ,\sin \beta \right),R\left( \cos \gamma ,\sin \gamma \right)$, whose orthocentre is $\left( 0,0 \right)$. Then, the value of $\cos \left( \alpha -\beta \right)+\cos \left( \beta -\gamma \right)+\cos \left( \gamma -\alpha \right)$ is

A. $\dfrac{-3}{2}$

B. $\dfrac{-1}{2}$

C. $\dfrac{1}{2}$

D. $\dfrac{3}{2}$

Answer

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Hint: The centroid of the triangle divides the line joining orthocentre and the circumcentre in the ratio \[2:1\].

The given vertices of the triangle are \[P(\cos \alpha ,\sin \alpha ),Q\left( \cos \beta ,\sin \beta \right)\] and \[R\left( \cos \gamma ,\sin \gamma \right)\].

Now, we will consider the circumcircle of the triangle.

To find the equation of the circumcircle , we will see the vertices of the triangle.

By observation , we can conclude that all the vertices of the triangle satisfy the equation \[{{x}^{2}}+{{y}^{2}}=1\]

So , the equation of the circumcircle will be \[{{x}^{2}}+{{y}^{2}}=1......(i)\]

Now , we know the general equation of circle with centre at \[(a,b)\] and radius \[r\] is given as

\[{{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}\]

Compared with equation\[(i)\], we can say that the centre of the circle represented by equation \[(i)\] is \[(0,0)\] and radius is \[1\] units.

So, the centre of the circumcircle is \[(0,0)\] and hence the circumcentre is \[(0,0)\].

Now, we know the centroid of the triangle divides the line joining orthocentre and the circumcentre in the ratio \[2:1\].

Now, in the question it is given that the orthocentre of \[\vartriangle ABC\] is \[(0,0)\] and from equation\[(i)\] we get that the circumcentre is also \[(0,0)\].

So , centroid is also \[(0,0)\].

We know , if a triangle has vertices \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\]and \[\left( {{x}_{3}},{{y}_{3}} \right)\] then , its centroid is given as

\[\left( \left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right),\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right) \right)\]

So , \[\left( \left( \dfrac{\cos \alpha +\cos \beta +\cos \gamma }{3} \right),\left( \dfrac{\sin \alpha +\sin \beta +\sin \gamma }{3} \right) \right)=\left( 0,0 \right)\]

Now,

\[\begin{align}

& \dfrac{\cos \alpha +\cos \beta +\cos \gamma }{3}=0 \\

& \Rightarrow \cos \alpha +\cos \beta +\cos \gamma =0 \\

\end{align}\]

\[\Rightarrow \cos \alpha +\cos \beta =-\cos \gamma \]

Now , we will square both sides .

On squaring on both the sides, we get

\[\Rightarrow {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +2\cos \alpha \cos \beta ={{\cos }^{2}}\gamma ......(ii)\]

Again , we have \[\dfrac{\sin \alpha +\sin \beta +\sin \gamma }{3}=0\]

\[\Rightarrow \sin \alpha +\sin \beta +\sin \gamma =0\]

\[\Rightarrow \sin \alpha +\sin \beta =-\sin \gamma \]

Now , we will square both sides .

On squaring on both the sides, we get

\[\Rightarrow {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +2\sin \alpha \sin \beta ={{\sin }^{2}}\gamma .......(iii)\]

Adding \[(ii)\]and \[(iii)\] we get ,\[\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)+\left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)+2\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)=\left( {{\cos }^{2}}\gamma +{{\sin }^{2}}\gamma \right)\]

We know , \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] and \[\cos A\cos B+\sin A\sin B=\cos (A-B)\] .

So , \[\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)+\left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)+2\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)=\left( {{\cos }^{2}}\gamma +{{\sin }^{2}}\gamma \right)\]can be written as \[1+1+2\left( \cos \left( \alpha -\beta \right) \right)=1\]

\[\Rightarrow 2\cos \left( \alpha -\beta \right)=-1\]

\[\Rightarrow \cos \left( \alpha -\beta \right)=\dfrac{-1}{2}........(iv)\]

Now , by symmetry we can say that

\[\cos \left( \beta -\gamma \right)=\dfrac{-1}{2}......(v)\]

And \[\cos \left( \gamma -\alpha \right)=\dfrac{-1}{2}.......(vi)\]

Adding \[(iv),(v)\]and \[(vi)\], we get

\[\begin{align}

& \cos \left( \alpha -\beta \right)+\cos \left( \beta -\alpha \right)+\cos \left( \gamma -\alpha \right)=\left( \dfrac{-1}{2} \right)+\left( \dfrac{-1}{2} \right)+\left( \dfrac{-1}{2} \right) \\

& \Rightarrow \cos \left( \alpha -\beta \right)+\cos \left( \beta -\alpha \right)+\cos \left( \gamma -\alpha \right)=\dfrac{-3}{2} \\

\end{align}\]

So , the value of $\cos \left( \alpha -\beta \right)+\cos \left( \beta -\gamma \right)+\cos \left( \gamma -\alpha \right)$ is equal to \[\dfrac{-3}{2}\].

Option (a) \[\dfrac{-3}{2}\] is correct answer

Note: Always remember that the centroid divides the line joining orthocentre and circumcentre in the ratio \[2:1\] and not \[1:2\]. Students often get confused and make mistakes.

The given vertices of the triangle are \[P(\cos \alpha ,\sin \alpha ),Q\left( \cos \beta ,\sin \beta \right)\] and \[R\left( \cos \gamma ,\sin \gamma \right)\].

Now, we will consider the circumcircle of the triangle.

To find the equation of the circumcircle , we will see the vertices of the triangle.

By observation , we can conclude that all the vertices of the triangle satisfy the equation \[{{x}^{2}}+{{y}^{2}}=1\]

So , the equation of the circumcircle will be \[{{x}^{2}}+{{y}^{2}}=1......(i)\]

Now , we know the general equation of circle with centre at \[(a,b)\] and radius \[r\] is given as

\[{{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}\]

Compared with equation\[(i)\], we can say that the centre of the circle represented by equation \[(i)\] is \[(0,0)\] and radius is \[1\] units.

So, the centre of the circumcircle is \[(0,0)\] and hence the circumcentre is \[(0,0)\].

Now, we know the centroid of the triangle divides the line joining orthocentre and the circumcentre in the ratio \[2:1\].

Now, in the question it is given that the orthocentre of \[\vartriangle ABC\] is \[(0,0)\] and from equation\[(i)\] we get that the circumcentre is also \[(0,0)\].

So , centroid is also \[(0,0)\].

We know , if a triangle has vertices \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\]and \[\left( {{x}_{3}},{{y}_{3}} \right)\] then , its centroid is given as

\[\left( \left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right),\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right) \right)\]

So , \[\left( \left( \dfrac{\cos \alpha +\cos \beta +\cos \gamma }{3} \right),\left( \dfrac{\sin \alpha +\sin \beta +\sin \gamma }{3} \right) \right)=\left( 0,0 \right)\]

Now,

\[\begin{align}

& \dfrac{\cos \alpha +\cos \beta +\cos \gamma }{3}=0 \\

& \Rightarrow \cos \alpha +\cos \beta +\cos \gamma =0 \\

\end{align}\]

\[\Rightarrow \cos \alpha +\cos \beta =-\cos \gamma \]

Now , we will square both sides .

On squaring on both the sides, we get

\[\Rightarrow {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +2\cos \alpha \cos \beta ={{\cos }^{2}}\gamma ......(ii)\]

Again , we have \[\dfrac{\sin \alpha +\sin \beta +\sin \gamma }{3}=0\]

\[\Rightarrow \sin \alpha +\sin \beta +\sin \gamma =0\]

\[\Rightarrow \sin \alpha +\sin \beta =-\sin \gamma \]

Now , we will square both sides .

On squaring on both the sides, we get

\[\Rightarrow {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +2\sin \alpha \sin \beta ={{\sin }^{2}}\gamma .......(iii)\]

Adding \[(ii)\]and \[(iii)\] we get ,\[\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)+\left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)+2\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)=\left( {{\cos }^{2}}\gamma +{{\sin }^{2}}\gamma \right)\]

We know , \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] and \[\cos A\cos B+\sin A\sin B=\cos (A-B)\] .

So , \[\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)+\left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)+2\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)=\left( {{\cos }^{2}}\gamma +{{\sin }^{2}}\gamma \right)\]can be written as \[1+1+2\left( \cos \left( \alpha -\beta \right) \right)=1\]

\[\Rightarrow 2\cos \left( \alpha -\beta \right)=-1\]

\[\Rightarrow \cos \left( \alpha -\beta \right)=\dfrac{-1}{2}........(iv)\]

Now , by symmetry we can say that

\[\cos \left( \beta -\gamma \right)=\dfrac{-1}{2}......(v)\]

And \[\cos \left( \gamma -\alpha \right)=\dfrac{-1}{2}.......(vi)\]

Adding \[(iv),(v)\]and \[(vi)\], we get

\[\begin{align}

& \cos \left( \alpha -\beta \right)+\cos \left( \beta -\alpha \right)+\cos \left( \gamma -\alpha \right)=\left( \dfrac{-1}{2} \right)+\left( \dfrac{-1}{2} \right)+\left( \dfrac{-1}{2} \right) \\

& \Rightarrow \cos \left( \alpha -\beta \right)+\cos \left( \beta -\alpha \right)+\cos \left( \gamma -\alpha \right)=\dfrac{-3}{2} \\

\end{align}\]

So , the value of $\cos \left( \alpha -\beta \right)+\cos \left( \beta -\gamma \right)+\cos \left( \gamma -\alpha \right)$ is equal to \[\dfrac{-3}{2}\].

Option (a) \[\dfrac{-3}{2}\] is correct answer

Note: Always remember that the centroid divides the line joining orthocentre and circumcentre in the ratio \[2:1\] and not \[1:2\]. Students often get confused and make mistakes.

Last updated date: 29th Sep 2023

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