Since my friends @Vishwak Srinivasan and @Satyajit Mohanty have already posted their methods , let me add one more for the sake of variety.

We know the friendly Sine rule:

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$

Applying Theorem on equal ratios , we have :

$\dfrac{a+b+c}{\sin A+ \sin B + \sin C} = 2R \\ \Rightarrow \dfrac{2s}{\sin A+ \sin B + \sin C} = 2R \\ \Rightarrow \sin A+ \sin B + \sin C = \dfrac{s}{R} \dots (1)$

We also have an identity (which can be proved only when $A+B+C=\pi$ ) :

$\sin A+ \sin B + \sin C = 4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2} \dots (2)$

Equating $(1),(2)$ , we get :

$4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}= \dfrac{s}{R} \\ \Rightarrow \cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}=\dfrac{s}{4R} = \dfrac{21}{4 \times 8.125} \\ \Rightarrow \cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}=\boxed{\dfrac{42}{65}}$

Thus, $x+y=42+65=\huge\boxed{107}$ .

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• $s$ is the semi-perimeter.

• $R$ is the circumradius.

• Upvote if you liked this :)

I am calling the expression to be found as $E$

In a triangle,

$\Delta = \dfrac{abc}{4R}$

$\cos\dfrac{A}{2} = \sqrt{\dfrac{s(s-a)}{bc}}$

The above formula is cyclic.

Hence:

$E =\cos\dfrac{A}{2} \cos\dfrac{B}{2} \cos\dfrac{C}{2} = \dfrac{s\sqrt{s(s-a)(s-b)(s-c)}}{\sqrt{a^2 b^2 c^2}} = \dfrac{s\Delta}{abc}$

$\Rightarrow E = \dfrac{s \frac{abc}{4R}}{abc} = \dfrac{s}{4R} = \dfrac{21}{32.5} = \dfrac{42}{65}$

$\Rightarrow x = 42 , y = 65 \Rightarrow x+y = 107$

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Note:

• $s$ is semi-perimeter

• $\Delta$ is Area of triangle

• $R$ is circumradius

• $a,b,c$ are the sides opposite to angles $A,B,C$ respectively.

• $\sqrt{s(s-a)(s-b)(s-c)} = \Delta$ by Heron's Formula

We'll use the four well known identities:

$r = 4Rsin(A/2)sin(B/2)sin(C/2)$ ; $a = 2Rsin(A)$ ; $R = \frac{abc}{4S}$ ; $r = \frac{S}{s}$ ,

Where $r$ = inradius, $R$ = circumradius, $S$ = Area, $s$ = semiperimeter; $A,B,C$ are angles and $a,b,c$ are the corresponding sides of the triangle.

Let $P = cos(A/2)cos(B/2)cos(C/2)$

=》 $8sin(A/2)sin(B/2)sin(C/2)P = sin(A)sin(B)sin(C)$

=》 $\frac{8rP}{4R} = \frac{abc}{(2R)^3}$

=》 $2rP = \frac{abc}{8R^2}$

=》 $P = \frac{abc}{16R^2r} = \frac{4S}{16Rr} = \frac{S}{4Rr} = \frac{s}{4R}$

We are given $s = 21, R = 8.125$ .

=》 $P = \frac{42}{65} = \frac{x}{y}$

=》 $x = 42, y = 65, x + y = \boxed{107}$

Use sine rule ie

a/sin(A)=b/sin(B)=c/sin(C)=2R

where a,b,c are lengths of triangle ABC and R is radius of circumcircle

also use conditional identites to prove sin(A)+sin(B)+sin(C)=4cos(A/2)cos(B/2)cos(C/2)