Probability = \(\frac{2}{5}\).

\(cot\theta =1\)
\(cot\theta =\cfrac { 1 }{ 3 } \)
\(\cfrac { adjacent }{ opposite } =\cfrac { 1 }{ 3 } \)
\(\sqrt { { 3 }^{ 2 }+1 } =\sqrt { 10 } \)
\(\cfrac { 3cos\theta -4sin\theta }{ 5sin\theta +4cos\theta } =\cfrac { 3\times \cfrac { 1 }{ \sqrt { 10 } } -4\times \cfrac { 3 }{ \sqrt { 10 } } }{ 5\times \cfrac { 3 }{ \sqrt { 10 } } +4\times \cfrac { 1 }{ \sqrt { 10 } } } =\cfrac { 3-12 }{ 15+4 } =\cfrac { -9 }{ 19 } \)

\(\left( \frac { 5(3)+3(5) }{ 5+3 } ,\frac { 5(1)+3(13) }{ 5+3 } \right) =\left( \frac { 15+15 }{ 8 } ,\frac { 5+39 }{ 8 } \right) =\left( \frac { 30 }{ 8 } ,\frac { 44 }{ 8 } \right) =\left( \frac { 15 }{ 4 } ,\frac { 11 }{ 2 } \right) \)

l= 20.5 m  b = 16 m  h = 8 m
∴ Capacity = Volume =I x b x h
= (20.5 \(\times\) 16 \(\times\)8) m³ = 2624 m³
1m³ = 1000 litres
∴ 2624 m³ = 2624000 litres.

(i) From the natural sines table
sin 85° 57' = 0.9975
\(\therefore\) \(\theta \) = 85° 57'
(ii) cos\(\theta \) = 0.6763
cos 47° 33' = 0.6762
\(\therefore\) \(\theta \) = 47° 33'
(iii) tan \(\theta \) = 0.0720
tan 4° 7' = 0.0720
\(\therefore\)\(\theta \) = 4°7'
(iv) cos \(\theta \) = 0.0410
cos 87° 45° = 0.0410
:\(\therefore\)\(\theta \) = 87°39'
(v) tan \(\theta \) = 7.5958
tan 82° 30' = 7.5958
\(\therefore\) \(\theta \) = 82° 30'

Let the monthly income of A and B be 3x and 4x respectively.
Let the monthly expenditure of A and B be 5y and 7y respectively.
\(\therefore\) 3x - 5y = 5000 --------- (1)
4x - 7y = 5000 --------- (2)
(1) \(\times\)4 \(\Rightarrow\) 12x - 20y = 20000
(2) \(\times\) 37 \(\Rightarrow\) \(\cfrac { 12-21y=15000 }{ 1y\quad =\quad 5000 } \)
Substitute y = 5000 in (1)
3x - 5 (5000) = 5000
3x - 25000 = 5000
3x = 5000 + 25000
3x = 30000
x = 10000
\(\therefore\) Monthly income of A is 3x = 3\(\times\)1 0000 = Rs. 30000
Monthly income of B is 4x = 4 \(\times\) 10000 = Rs. 40000

(i) \(sinB=\cfrac { 12 }{ 13 } \)
(ii) \(secB=\cfrac { 1 }{ cosB } \)
=  \(\cfrac { 1 }{ 5/3 } =\cfrac { 13 }{ 5 } \)
(iii) \(cotB=\cfrac { 1 }{ tanB } =\cfrac { 1 }{ 12/5 } =\cfrac { 5 }{ 2 } \)
(iv) 

(v) \(\tan C=\cfrac { 12 }{ 16 } =\cfrac { 3 }{ 4 } \)
(vi)  \(cosecC=\cfrac { 1 }{ sinC } =\cfrac { 1 }{ 12/20 } =\cfrac { 20 }{ 12 } =\cfrac { 5 }{ 3 } \)

By the pythagoras theorem,
\(AD=\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } \)
= \(\sqrt { 169-25 } \)
= \(\sqrt { 144 } =12\)
AC =\(\sqrt { { 12 }^{ 2 }+{ 16 }^{ 2 } } \)
= \(\sqrt { { 144 }+{ 256 } } \)
= \(\sqrt { 400 } =20\) 

(i) Let \(x=0.\overline { 3 } \)= 0.3333   (1)
(Here period of decimal is 1, multiply equation (1) by 10)
10x = 3.3333....    (2)
(2) - (1) : 9x = 3  or \(x=\frac { 1 }{ 3 } \)
(ii) Let \(x=2.\overline { 124 } \) = 2.124124124…            (1)
(Here period of decimal is 3, multiply equation (1) by 1000)
1000 x = 2124.124124124… (2)
(2)–(1): 999 x = 2122  \(x=\frac { 2122 }{ 999 } \)
(iii) Let x = \(0.4\overline { 5 } \) = 0.45555…              (1)
(Here the repeating decimal digit is 5, which is the second digit after the decimal point, multiply equation (1) by 10)
10 x = 4.5555…  (2)
(Now period of decimal is 1, multiply equation (2) by 10)
100 x = 45.5555…              (3)
(3) – (2): 90 x = 41 or \(x=\frac { 41 }{ 90 } \)
(iv) Let x = \(0.5\overline { 68 } \)=  0.5686868…  (1)
(Here the repeating decimal digit is 68, which is the second digit after the decimal point, so multiply equation (1) by 10)
10 x = 5.686868…                (2)
(Now period of decimal is 2, multiply equation (2) by 100)
1000 x = 568.686868… (3)
3) – (2): 990 x = 563 or \(x=\frac { 563 }{ 990 } \)

x⁴ - 9x² = x² (x² - 9) = x² (x² - 32) = x² (x - 3)(x + 3)

Distance of the chord from the centre
= \(\sqrt { { 15 }^{ 2 }-{ 12 }^{ 2 } } \)
= \(\sqrt { 225-144 } \)
= \(\sqrt { 89 } \)
= 9 cm

In the figure\(\angle \)POQ = 100⁰
\(\angle \)PQR = 30⁰
\(\angle \)PQR = \(\cfrac { 1 }{ 2 } \)\(\angle \)PQR = \(\cfrac { 1 }{ 2 } \)\(\times\)100 = 50⁰
In\(\triangle \)OPQ = \(\angle \)OQP =\(\angle \)POQ = 180⁰
2\(\angle \)OPQ = 180⁰
2\(\angle \) OPQ = 80⁰
\(\therefore \) \(\angle \)OPQ = 40⁰ 
In \(\triangle \)PRQ,
\(\angle \)R+\(\angle \)P+\(\angle \)Q = 180⁰
50⁰ + (40 + x) + 30⁰ = 180⁰
(40 + x)⁰ = 180⁰ - 80 = 100⁰
x⁰ = 100⁰ - 40⁰ = 60⁰
\(\therefore \) \(\angle \)RPO = x = 60⁰

\(m^2+\frac{1}{m^2}-23=(m+\frac{1}{m})^2-2-23=(m+\frac{1}{m})^2-25\)
=\((m+\frac{1}{m})^2-5^2\)
\(\left( m+\frac { 1 }{ m } +5 \right) \left( m+\frac { 1 }{ m } -5 \right) \)

(m + 5)(m - 8)   [we have (x + a)(x - b) = x²- (a - b)x - ab]
(m + 5)(m - 8) = m²+(5 - 8)m - (5)(8)   put [x = m, a = 5, b = 8]
= m²- 3m - 40

\(\left( \frac { 64 }{ 125 } \right) ^{ \frac { -2 }{ 3 } }=(\frac{125}{64})^{\frac{2}{3}}=((\frac{5}{4})^3)^{\frac{2}{3}}=(\frac{5}{4})^{ ̶3̶\times\frac{2}{ ̶3̶}}=(\frac{5}{4})^2=\frac{25}{16}\)

\({17\over 200}={17\over 2^3\times 5^2}\)
\(\therefore {17\over 200}\) has a terminating decimal expansion.

A\(\cup \)B = {a, b, c, e, u} \(\cup \) {a, e, i, o, u}
= {a, b, c, e, i, o, u}
A\(\cap\)B = {a, b, c, e, u} \(\cap\) {a, e, i, o, u}
= {a, e, u}
A-B = {a, b, c, e, u} - {a, e, i, o, u}
= {b, c}
B-A = {a, e, i, o, u} - {a, b, c, e, u}
= {i, o}

{P, A, R, L, E, O, G, M}

Let f(x) = x² – 9
Then, f(–3) = (–3)² – 9 = 9 – 9 = 0
f(+3) = 3² – 9 = 9 – 9 = 0
\(\therefore\) -3 and 3 are zeros of the polynomial x² – 9

It is a trapezium

It is a parallelogram

A′

A – B = {–3, -2, 1, 4} – {0, 1, 2, 4} = { -3, -2}
B – A = {0, 1, 2, 4} – {–3, -2, 1, 4} = { 0, 2}.

PRINCIPAL
B = {P, R, I, N, C, A, L}

(i) AUB

(ii) A∩B

(iii) (A∩B)′

(iv) (B–A)'

(v) A′UB′

(vi) A′∩B′


(vii) From the venn diagram (iii) and (v), We have
\((A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}\)
Now \(\mathrm{B} \cap \mathrm{C}=\left\{\frac{1}{4}, 2, \frac{5}{2}\right\}\)
\(A \cap(B \cap C)=\left\{\frac{1}{4}, 2\right\}\)............(1)
Then \(A \cap B=\left\{0, \frac{1}{4}, \frac{3}{4}, 2\right\}\)
\((A \cap B) \cap C=\left\{\frac{1}{4}, 2\right\}\).............(2)
From (1) and (2), it is verified that
\((A \cap B) \cap C=A \cap(B \cap C)\)