Answer
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Hint: Equate $\dfrac{s-x}{4}=\dfrac{s-y}{3}=\dfrac{s-z}{2}$to a constant k to get values of x, y, z in terms of 5.
Complete step-by-step answer:
Get the radius of incircle using the area of the circle as $\pi {{r}^{2}}$. Use the relation $r=\dfrac{\Delta }{s}$ to get the value of area i.e. $\Delta $ in terms of ‘s’. calculate $\Delta $in terms of ‘s’ by using heron’s formula. It is given as
$\Delta =\sqrt{s\left( s-x \right)\left( s-y \right)\left( s-z \right)}$
Where, $s=\dfrac{x+y+z}{2}$, x, y, z are sides of the triangle.
Use relation $r=4R\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2}$ to get value of $\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2}$.
And use the cosine formula as well; it is given as $\cos \theta =\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$
Where, $\theta $ is the angle between ‘a’ ‘b’, and a, b, c are representing the sides of a triangle.
Here, we are given a triangle xyz, with sides of lengths x, y, z opposite to the angles x, y, z respectively, and hence, we need to verify the given options, if following equations are given as
$2s = x + y + z ……………………(i)$
$\dfrac{s-x}{4}=\dfrac{s-y}{3}=\dfrac{s-z}{2}$……………….(ii)
Area of in circle =\[\dfrac{8\pi }{3}\]…………………(iii)
Let us equate all the equal fractions of equation (ii) to a constant value ‘k’. so, we get
\[\dfrac{s-x}{4}=k,\dfrac{s-y}{3}=k,\dfrac{s-z}{2}=k\]
\[s-x=4k\] ………………(iv)
\[s-y=3k\] ………………(v)
\[s-z=2k\] …………….(vi) \[\]
Now, we can add above three equations as
(s - x) + (s – y) + (s – z) = 4k + 3k + 2k
3s – (x + y + z) = 9k
Now, we can replace x + y + z in the above equation by ‘2s’ from the equation (i). so, we get
3s – 2s = 9k
s = 9k,
$k=\dfrac{s}{9}$
Now, we can get values of x, y, z from the equations (iv), (v), (vi) as
$\begin{align}
& s-x=\dfrac{4s}{9} \\
& x=s-\dfrac{4s}{9}=\dfrac{5s}{9} \\
& x=\dfrac{5s}{9} \\
\end{align}$
Similarly, we get value of ‘y’ as
$\begin{align}
& s-y=\dfrac{3s}{9} \\
& y=s-\dfrac{3s}{9}=\dfrac{6s}{9}=\dfrac{2s}{3} \\
& y=\dfrac{2s}{3} \\
\end{align}$
And, we get value of ‘z’ as
$\begin{align}
& s-z=\dfrac{2s}{9} \\
& z=s-\dfrac{2s}{9}=\dfrac{7s}{9} \\
& z=\dfrac{7s}{9} \\
\end{align}$
Hence, we get value of (x, y, z) as
$\left( \dfrac{5s}{9},\dfrac{2s}{3},\dfrac{7s}{9} \right)$
Now, we know the area of a circle with radius R is given as $\pi {{r}^{2}}$.
So, let radius of incircle be ‘r’
So, we get area of in circle as = $\pi {{r}^{2}}$
We are given the area of the circle as $\dfrac{8\pi }{3}$ from the equation (iii). So, we get
$\begin{align}
& \pi {{r}^{2}}=\dfrac{8\pi }{3} \\
& {{r}^{2}}=\dfrac{8}{3} \\
\end{align}$
$r=\sqrt{\dfrac{8}{3}}$…………….(vii)
Now, we know the relation among
$s=\dfrac{x+y+z}{2}$, radius of in circle ‘r’ and area of the triangle $'\Delta '$ is
$\dfrac{\Delta }{s}=r$ ……………………..(viii)
So, we can get relation from equations (vii) and (viii) as
$\begin{align}
& \dfrac{\Delta }{s}=\sqrt{\dfrac{8}{3}} \\
& \Delta =s\sqrt{\dfrac{8}{3}} \\
\end{align}$
On squaring both sides, we get
\[{{\Delta }^{2}}={{s}^{2}}\dfrac{8}{3}=\dfrac{8}{3}{{s}^{2}}\]
\[{{\Delta }^{2}}=\dfrac{8{{s}^{2}}}{3}\]……………(ix)
Now, we know area of a triangle by heron’s formula can be given as
\[\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]……………….. (x)
Where, a, b, c are sides of the triangle and $s=\dfrac{a+b+c}{2}$.
So, we can get area of triangle given here as
$\Delta =\sqrt{s\left( s-x \right)\left( s-y \right)\left( s-z \right)}$
Now, we can put values of (x, y, z) to the above equation as $\left( \dfrac{5s}{9},\dfrac{2s}{3},\dfrac{7s}{9} \right)$.
So, we get
$\begin{align}
& \Delta =\sqrt{s\left( s-\dfrac{5s}{9} \right)\left( s-\dfrac{2s}{3} \right)\left( s-\dfrac{7s}{9} \right)} \\
& \Delta =\sqrt{\left( s \right)\left( \dfrac{4s}{9} \right)\left( \dfrac{s}{3} \right)\left( \dfrac{2s}{9} \right)} \\
& \Delta =\sqrt{\dfrac{8{{s}^{4}}}{{{243}}}} \\
\end{align}$
Squaring both sides of the above equation, we get
${{\Delta }^{2}}=\dfrac{8{{s}^{4}}}{243}$……………….(xi)
Now, from equation (ix) and (xi), we get
$\begin{align}
& \dfrac{8{{s}^{4}}}{3\times 81}=\dfrac{8}{3}{{s}^{2}} \\
& {{s}^{2}}=81 \\
\end{align}$
s = 9
So, we get value of ‘s’ as
s = 9…………….(xii)
hence, from equation (ix), we get
${{\Delta }^{2}}=\dfrac{8}{3}{{s}^{2}}=\dfrac{8}{3}\times 9\times 9=27\times 8$
Taking square root to both the sides of the equation, we get
$\begin{align}
& \Delta =3\times 2\sqrt{3\times 2}=6\sqrt{6} \\
& \Delta =6\sqrt{6} \\
\end{align}$
Hence, the area of triangle xyz$\to \text{6}\sqrt{\text{6}}\text{ square units}$.
So, option (a) is the correct answer.
We know the relation sides of a triangle x, y, z and radius of circumcircle ‘R’ and area of triangle $'\Delta '$ can be given as
$R=\dfrac{xyz}{4\Delta }$
So, we can put values of x, y, z and $\Delta $ to the above equation so, we get
$\begin{align}
& R=\dfrac{\dfrac{5s}{9}\times \dfrac{2s}{3}\times \dfrac{7s}{9}}{4\times 6\sqrt{6}} \\
& R=\dfrac{70{{s}^{3}}}{9\times 3\times 9\times 6\sqrt{6}\times 4} \\
\end{align}$
We can put the value of s as 9. So, we get
$\begin{align}
& R=\dfrac{70\times 9\times 9\times 9}{9\times 3\times 9\times 6\sqrt{6}\times 4} \\
& R=\dfrac{35}{4\sqrt{6}} \\
\end{align}$
Multiply and divide the above equation by $\sqrt{6}$, we get
$\begin{align}
& R=\dfrac{35}{4\sqrt{6}}\times \dfrac{\sqrt{6}}{\sqrt{6}}=\dfrac{35\sqrt{6}}{24} \\
& R=\dfrac{35\sqrt{6}}{24} \\
\end{align}$
Hence, option (b) i.e. Radius of circumcircle of the triangle xyz is $\dfrac{35\sqrt{6}}{6}$ is incorrect as radius of circumcircle is $\dfrac{35\sqrt{6}}{24}$.
Now, we know the relation
$r=4R\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2}$ ………….(xiii)
Where, x, y, z are angles of the triangle r and R are radius of incircle and circumcircle respectively, so, we can put value of r and R as $\sqrt{\dfrac{8}{3}},\dfrac{35}{4\sqrt{6}}$to the equation (xiii) to get the value of $\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2}.$
So, we get
$\begin{align}
& \sqrt{\dfrac{8}{3}}=\dfrac{35}{4\sqrt{6}}\times 4\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2} \\
& \dfrac{\sqrt{6}}{35}\times \dfrac{\sqrt{8}}{\sqrt{3}}=\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2} \\
& \dfrac{\sqrt{16}}{35}=\dfrac{4}{35}=\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2} \\
\end{align}$
So, we get
$\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2}=\dfrac{4}{35}$
Hence, option (c) is the correct answer.
Now, we can get the value of ${{\sin }^{2}}\left( \dfrac{x+y}{2} \right)$ by the following way. Let the value of ${{\sin }^{2}}\dfrac{x+y}{2}\to 'M'$. $M={{\sin }^{2}}\left( \dfrac{x+y}{2} \right)$ …………………(xiv)
As we know, the sum of interior angles of a triangle is ${{180}^{\circ }}$. So, we get
x + y + z = ${{180}^{\circ }}$.
x + y = 180 – z
so, we get equation (xiv) as
$\begin{align}
& M={{\sin }^{2}}\left( \dfrac{180-z}{2} \right) \\
& M={{\sin }^{2}}\left( 90-\dfrac{z}{2} \right) \\
& M={{\left( \sin \left( 90-\dfrac{z}{2} \right) \right)}^{2}} \\
\end{align}$
We know
$\sin \left( 90-\theta \right)=\cos \theta $
So, we get value of M as
$\begin{align}
& M={{\cos }^{2}}\dfrac{z}{2} \\
& M={{\cos }^{2}}\dfrac{z}{2} \\
\end{align}$
We know the trigonometric identity of $\cos 2\theta $ is given as
$\begin{align}
& \cos 2\theta =2{{\cos }^{2}}\theta -1. \\
& {{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2} \\
\end{align}$
Now, we can put $\theta =\dfrac{z}{2}$to the above equation, so, we get
${{\cos }^{2}}\left( \dfrac{z}{2} \right)=\dfrac{1+\cos z}{2}$
So, we get value of M as
$M=\dfrac{1+\cos z}{2}$ ……………….(xv)
Now, we know the cosine formula is given as
$\cos \theta =\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$
Where $\theta $ is the angle between ‘b’, ‘c’, and a, b, c are the sides of a triangle.
So, we can write value of $\cos z$ from the above equation and using diagram as
$\operatorname{cosz}=\dfrac{{{x}^{2}}+{{y}^{2}}-{{z}^{2}}}{2xy}$
Now, we can put value of (x, y, z) $\to \left( \dfrac{5s}{9},\dfrac{2s}{3},\dfrac{7s}{9} \right)$ to the above equation. So, we get
$\begin{align}
& \cos z=\dfrac{{{\left( \dfrac{5s}{9} \right)}^{2}}+{{\left( \dfrac{2s}{3} \right)}^{2}}-{{\left( \dfrac{7s}{9} \right)}^{2}}}{2\times \dfrac{5s}{9}\times \dfrac{2s}{3}} \\
& \cos z=\dfrac{\dfrac{25{{s}^{2}}}{81}+\dfrac{4{{s}^{2}}}{9}-\dfrac{49{{s}^{2}}}{81}}{\dfrac{20{{s}^{2}}\times 3}{81}} \\
& \cos z=\dfrac{\dfrac{25{{s}^{2}}+36{{s}^{2}}-49{{s}^{2}}}{81}}{\dfrac{60{{s}^{2}}}{81}} \\
& \cos z=\dfrac{{{s}^{2}}\left[ 61-49 \right]}{60{{s}^{2}}}=\dfrac{12}{60} \\
& \cos z=\dfrac{1}{5} \\
\end{align}$
Hence, we can get value of M as
$\begin{align}
& M=\dfrac{1+\dfrac{1}{5}}{2} \\
& M=\dfrac{6}{5\times 2}=\dfrac{3}{5} \\
& M=\dfrac{3}{5} \\
\end{align}$
Hence, value of ${{\sin }^{2}}\left( \dfrac{x+2}{2} \right)\to \dfrac{3}{5}$
So, option (d) is the correct answer.
Note: Identities play an important role to solve these kinds of problems. Relations among the terms involved in the question makes the solution flexible and will take less time.
One may involve the sine rule for the calculation of ${{\sin }^{2}}\left( \dfrac{x+y}{2} \right)\to \left( \dfrac{1+\cos z}{2} \right)$. It is given as
$\dfrac{\sin x}{x}=\dfrac{\sin y}{y}=\dfrac{\sin z}{z}=\dfrac{1}{2R}$
Where, we can get $\sin z$, by putting the value of z and R.
And hence, calculate $\cos z$ by the equation $\cos z=\sqrt{1-{{\sin }^{2}}z}$.
Complete step-by-step answer:
Get the radius of incircle using the area of the circle as $\pi {{r}^{2}}$. Use the relation $r=\dfrac{\Delta }{s}$ to get the value of area i.e. $\Delta $ in terms of ‘s’. calculate $\Delta $in terms of ‘s’ by using heron’s formula. It is given as
$\Delta =\sqrt{s\left( s-x \right)\left( s-y \right)\left( s-z \right)}$
Where, $s=\dfrac{x+y+z}{2}$, x, y, z are sides of the triangle.
Use relation $r=4R\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2}$ to get value of $\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2}$.
And use the cosine formula as well; it is given as $\cos \theta =\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$
Where, $\theta $ is the angle between ‘a’ ‘b’, and a, b, c are representing the sides of a triangle.
Here, we are given a triangle xyz, with sides of lengths x, y, z opposite to the angles x, y, z respectively, and hence, we need to verify the given options, if following equations are given as
$2s = x + y + z ……………………(i)$
$\dfrac{s-x}{4}=\dfrac{s-y}{3}=\dfrac{s-z}{2}$……………….(ii)
Area of in circle =\[\dfrac{8\pi }{3}\]…………………(iii)
Let us equate all the equal fractions of equation (ii) to a constant value ‘k’. so, we get
\[\dfrac{s-x}{4}=k,\dfrac{s-y}{3}=k,\dfrac{s-z}{2}=k\]
\[s-x=4k\] ………………(iv)
\[s-y=3k\] ………………(v)
\[s-z=2k\] …………….(vi) \[\]
Now, we can add above three equations as
(s - x) + (s – y) + (s – z) = 4k + 3k + 2k
3s – (x + y + z) = 9k
Now, we can replace x + y + z in the above equation by ‘2s’ from the equation (i). so, we get
3s – 2s = 9k
s = 9k,
$k=\dfrac{s}{9}$
Now, we can get values of x, y, z from the equations (iv), (v), (vi) as
$\begin{align}
& s-x=\dfrac{4s}{9} \\
& x=s-\dfrac{4s}{9}=\dfrac{5s}{9} \\
& x=\dfrac{5s}{9} \\
\end{align}$
Similarly, we get value of ‘y’ as
$\begin{align}
& s-y=\dfrac{3s}{9} \\
& y=s-\dfrac{3s}{9}=\dfrac{6s}{9}=\dfrac{2s}{3} \\
& y=\dfrac{2s}{3} \\
\end{align}$
And, we get value of ‘z’ as
$\begin{align}
& s-z=\dfrac{2s}{9} \\
& z=s-\dfrac{2s}{9}=\dfrac{7s}{9} \\
& z=\dfrac{7s}{9} \\
\end{align}$
Hence, we get value of (x, y, z) as
$\left( \dfrac{5s}{9},\dfrac{2s}{3},\dfrac{7s}{9} \right)$
Now, we know the area of a circle with radius R is given as $\pi {{r}^{2}}$.
So, let radius of incircle be ‘r’
So, we get area of in circle as = $\pi {{r}^{2}}$
We are given the area of the circle as $\dfrac{8\pi }{3}$ from the equation (iii). So, we get
$\begin{align}
& \pi {{r}^{2}}=\dfrac{8\pi }{3} \\
& {{r}^{2}}=\dfrac{8}{3} \\
\end{align}$
$r=\sqrt{\dfrac{8}{3}}$…………….(vii)
Now, we know the relation among
$s=\dfrac{x+y+z}{2}$, radius of in circle ‘r’ and area of the triangle $'\Delta '$ is
$\dfrac{\Delta }{s}=r$ ……………………..(viii)
So, we can get relation from equations (vii) and (viii) as
$\begin{align}
& \dfrac{\Delta }{s}=\sqrt{\dfrac{8}{3}} \\
& \Delta =s\sqrt{\dfrac{8}{3}} \\
\end{align}$
On squaring both sides, we get
\[{{\Delta }^{2}}={{s}^{2}}\dfrac{8}{3}=\dfrac{8}{3}{{s}^{2}}\]
\[{{\Delta }^{2}}=\dfrac{8{{s}^{2}}}{3}\]……………(ix)
Now, we know area of a triangle by heron’s formula can be given as
\[\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]……………….. (x)
Where, a, b, c are sides of the triangle and $s=\dfrac{a+b+c}{2}$.
So, we can get area of triangle given here as
$\Delta =\sqrt{s\left( s-x \right)\left( s-y \right)\left( s-z \right)}$
Now, we can put values of (x, y, z) to the above equation as $\left( \dfrac{5s}{9},\dfrac{2s}{3},\dfrac{7s}{9} \right)$.
So, we get
$\begin{align}
& \Delta =\sqrt{s\left( s-\dfrac{5s}{9} \right)\left( s-\dfrac{2s}{3} \right)\left( s-\dfrac{7s}{9} \right)} \\
& \Delta =\sqrt{\left( s \right)\left( \dfrac{4s}{9} \right)\left( \dfrac{s}{3} \right)\left( \dfrac{2s}{9} \right)} \\
& \Delta =\sqrt{\dfrac{8{{s}^{4}}}{{{243}}}} \\
\end{align}$
Squaring both sides of the above equation, we get
${{\Delta }^{2}}=\dfrac{8{{s}^{4}}}{243}$……………….(xi)
Now, from equation (ix) and (xi), we get
$\begin{align}
& \dfrac{8{{s}^{4}}}{3\times 81}=\dfrac{8}{3}{{s}^{2}} \\
& {{s}^{2}}=81 \\
\end{align}$
s = 9
So, we get value of ‘s’ as
s = 9…………….(xii)
hence, from equation (ix), we get
${{\Delta }^{2}}=\dfrac{8}{3}{{s}^{2}}=\dfrac{8}{3}\times 9\times 9=27\times 8$
Taking square root to both the sides of the equation, we get
$\begin{align}
& \Delta =3\times 2\sqrt{3\times 2}=6\sqrt{6} \\
& \Delta =6\sqrt{6} \\
\end{align}$
Hence, the area of triangle xyz$\to \text{6}\sqrt{\text{6}}\text{ square units}$.
So, option (a) is the correct answer.
We know the relation sides of a triangle x, y, z and radius of circumcircle ‘R’ and area of triangle $'\Delta '$ can be given as
$R=\dfrac{xyz}{4\Delta }$
So, we can put values of x, y, z and $\Delta $ to the above equation so, we get
$\begin{align}
& R=\dfrac{\dfrac{5s}{9}\times \dfrac{2s}{3}\times \dfrac{7s}{9}}{4\times 6\sqrt{6}} \\
& R=\dfrac{70{{s}^{3}}}{9\times 3\times 9\times 6\sqrt{6}\times 4} \\
\end{align}$
We can put the value of s as 9. So, we get
$\begin{align}
& R=\dfrac{70\times 9\times 9\times 9}{9\times 3\times 9\times 6\sqrt{6}\times 4} \\
& R=\dfrac{35}{4\sqrt{6}} \\
\end{align}$
Multiply and divide the above equation by $\sqrt{6}$, we get
$\begin{align}
& R=\dfrac{35}{4\sqrt{6}}\times \dfrac{\sqrt{6}}{\sqrt{6}}=\dfrac{35\sqrt{6}}{24} \\
& R=\dfrac{35\sqrt{6}}{24} \\
\end{align}$
Hence, option (b) i.e. Radius of circumcircle of the triangle xyz is $\dfrac{35\sqrt{6}}{6}$ is incorrect as radius of circumcircle is $\dfrac{35\sqrt{6}}{24}$.
Now, we know the relation
$r=4R\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2}$ ………….(xiii)
Where, x, y, z are angles of the triangle r and R are radius of incircle and circumcircle respectively, so, we can put value of r and R as $\sqrt{\dfrac{8}{3}},\dfrac{35}{4\sqrt{6}}$to the equation (xiii) to get the value of $\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2}.$
So, we get
$\begin{align}
& \sqrt{\dfrac{8}{3}}=\dfrac{35}{4\sqrt{6}}\times 4\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2} \\
& \dfrac{\sqrt{6}}{35}\times \dfrac{\sqrt{8}}{\sqrt{3}}=\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2} \\
& \dfrac{\sqrt{16}}{35}=\dfrac{4}{35}=\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2} \\
\end{align}$
So, we get
$\sin \dfrac{x}{2}\sin \dfrac{y}{2}\sin \dfrac{z}{2}=\dfrac{4}{35}$
Hence, option (c) is the correct answer.
Now, we can get the value of ${{\sin }^{2}}\left( \dfrac{x+y}{2} \right)$ by the following way. Let the value of ${{\sin }^{2}}\dfrac{x+y}{2}\to 'M'$. $M={{\sin }^{2}}\left( \dfrac{x+y}{2} \right)$ …………………(xiv)
As we know, the sum of interior angles of a triangle is ${{180}^{\circ }}$. So, we get
x + y + z = ${{180}^{\circ }}$.
x + y = 180 – z
so, we get equation (xiv) as
$\begin{align}
& M={{\sin }^{2}}\left( \dfrac{180-z}{2} \right) \\
& M={{\sin }^{2}}\left( 90-\dfrac{z}{2} \right) \\
& M={{\left( \sin \left( 90-\dfrac{z}{2} \right) \right)}^{2}} \\
\end{align}$
We know
$\sin \left( 90-\theta \right)=\cos \theta $
So, we get value of M as
$\begin{align}
& M={{\cos }^{2}}\dfrac{z}{2} \\
& M={{\cos }^{2}}\dfrac{z}{2} \\
\end{align}$
We know the trigonometric identity of $\cos 2\theta $ is given as
$\begin{align}
& \cos 2\theta =2{{\cos }^{2}}\theta -1. \\
& {{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2} \\
\end{align}$
Now, we can put $\theta =\dfrac{z}{2}$to the above equation, so, we get
${{\cos }^{2}}\left( \dfrac{z}{2} \right)=\dfrac{1+\cos z}{2}$
So, we get value of M as
$M=\dfrac{1+\cos z}{2}$ ……………….(xv)
Now, we know the cosine formula is given as
$\cos \theta =\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$
Where $\theta $ is the angle between ‘b’, ‘c’, and a, b, c are the sides of a triangle.
So, we can write value of $\cos z$ from the above equation and using diagram as
$\operatorname{cosz}=\dfrac{{{x}^{2}}+{{y}^{2}}-{{z}^{2}}}{2xy}$
Now, we can put value of (x, y, z) $\to \left( \dfrac{5s}{9},\dfrac{2s}{3},\dfrac{7s}{9} \right)$ to the above equation. So, we get
$\begin{align}
& \cos z=\dfrac{{{\left( \dfrac{5s}{9} \right)}^{2}}+{{\left( \dfrac{2s}{3} \right)}^{2}}-{{\left( \dfrac{7s}{9} \right)}^{2}}}{2\times \dfrac{5s}{9}\times \dfrac{2s}{3}} \\
& \cos z=\dfrac{\dfrac{25{{s}^{2}}}{81}+\dfrac{4{{s}^{2}}}{9}-\dfrac{49{{s}^{2}}}{81}}{\dfrac{20{{s}^{2}}\times 3}{81}} \\
& \cos z=\dfrac{\dfrac{25{{s}^{2}}+36{{s}^{2}}-49{{s}^{2}}}{81}}{\dfrac{60{{s}^{2}}}{81}} \\
& \cos z=\dfrac{{{s}^{2}}\left[ 61-49 \right]}{60{{s}^{2}}}=\dfrac{12}{60} \\
& \cos z=\dfrac{1}{5} \\
\end{align}$
Hence, we can get value of M as
$\begin{align}
& M=\dfrac{1+\dfrac{1}{5}}{2} \\
& M=\dfrac{6}{5\times 2}=\dfrac{3}{5} \\
& M=\dfrac{3}{5} \\
\end{align}$
Hence, value of ${{\sin }^{2}}\left( \dfrac{x+2}{2} \right)\to \dfrac{3}{5}$
So, option (d) is the correct answer.
Note: Identities play an important role to solve these kinds of problems. Relations among the terms involved in the question makes the solution flexible and will take less time.
One may involve the sine rule for the calculation of ${{\sin }^{2}}\left( \dfrac{x+y}{2} \right)\to \left( \dfrac{1+\cos z}{2} \right)$. It is given as
$\dfrac{\sin x}{x}=\dfrac{\sin y}{y}=\dfrac{\sin z}{z}=\dfrac{1}{2R}$
Where, we can get $\sin z$, by putting the value of z and R.
And hence, calculate $\cos z$ by the equation $\cos z=\sqrt{1-{{\sin }^{2}}z}$.
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