101/144, 47/72

-7/20,-3/10,-1/4

610/33

\(5+2\sqrt { 3 } -3\sqrt { 2 } \)

(i) 57
(ii) \(\frac { 1 }{ 3129 } \)

Quotient = 3x - 2
Remainder = 1

11130

(2,3), (2,0), (0,3)

\(90^{ 0 }\)

\((a)\Delta DAB,DCAB\)
\(\\ (b)\Delta DAB,||gm\ ABCD\)

p(x) = x³ - 3x² - 2x + 6
Then, p(\(\sqrt { 2 } \)) = (\(\sqrt { 2 } \))³ -3(\(\sqrt { 2 } \))² - 2(\(\sqrt { 2 } \)) + 6
=2\(\sqrt { 2 } \) - 6 - 2\(\sqrt { 2 } \) + 6
 = 0

(i) Infinite, if they are collinear.
(ii) Only one, if they are non-collinear points.

\(\left( x+\frac { 1 }{ x } \right) \left( x-\frac { 1 }{ x } \right) \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) \left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } \right) \)
\(=\left( { x }^{ 2 }-\frac { 1 }{ { x }^{ 2 } } \right) \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) \left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } \right) \)
as (a+b)(a-b)=a²-b²
\(=\left( { x }^{ 4 }-\frac { 1 }{ { x }^{ 4 } } \right) \left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } \right) \)
\(=\left( { x }^{ 8 }-\frac { 1 }{ { x }^{ 8 } } \right) \)

Given, \(\angle COA=82^o\)
\(\Rightarrow \angle COB+\angle BOA=82^o\)
\(\Rightarrow \angle COB+35^o=82^o(\because \angle BOA=35^o)\)
\(\Rightarrow \angle COB=82^o-35^o=47^o\)
Similarly, \(\angle DOB=\angle DOC+\angle COB\)
\(\Rightarrow 87^o=\angle DOC+47^o\)
\(\Rightarrow \angle DOC=87^o-47^o=40^o\)

\(\angle BMN=\angle CNM(alternate \angle 's)\)
\(\frac{1}{2}\angle BMN=\frac{1}{2}\angle CNM\)
x=y
(But x and y are alternate angles)
MP || NQ

3z-42^o=2z+13^o (Alternate interior angles)
z=42^o+13^o
z=55^o
\(\angle DNM=2z+13^o=110^o+13^o\)
=123^o
\(\angle CNM=180^o-\angle DNM\)(Linear pair)
=180^o-123^o=57^o

Let the probability of winning a game = p
and probability of losing a game = q
we know that p + q = 1...(i)
According to questions,
\(p=2q-\frac{1}{3}\)
\(\Rightarrow \) 6q - 3p = 1...(ii)
On solving (i) and (ii), we get
\(q=\frac{4}{9}\ and\ p=\frac{5}{9}\)
\(\therefore \) Probability of winning game \(=\frac{5}{9}\)

Since sum of the opposite pairs of angles in a cyclic quadrilateral is 180°.
Hence ㄥB+ㄥD=180°
ㄥB=180°-70°
=110°
Again, AB II CD and AD is its transversal, so
ㄥA+ㄥD=180°
ㄥA=180°-70°
=110°
and ㄥA+ㄥC=180°
⇒ 110°+ㄥC=180°
ㄥC=180°-110°
=70°