l = [angle / 360o] * 2pr                    p=pie

= [120 / 360] * 2 (22 / 7) * (2.1)

= [0.333] * 13.195

= 4.4cm

Area of the sector = ½ r² angle

= (0.5) * (2.1)2 * (120o)

= 264.6 cm2

To find the height of the tower CD

Draw BE parallel to AC and DQ.

Lines DQ and BE are parallel, BD is the transversal.

DBE = QDB [alternate angles]

DBE = 30o

Lines DQ and AC are parallel, AD is the transversal.

DAC = QDA [alternate angles]

DAC = 60o

AC and BE are parallel lines.

AC = BE

Similarly, AB and CE are also parallel lines.

CD = AB

CD = 50m

Since DC is perpendicular to AC, ?DEB = ?DCA = 90o.

tan 30o = DE / BE

1 / v3 = DE / BE

BE = v3 DE — (1)

tan 60o = DC / AC

3 = DC / AC

AC = DC / v3 — (2)

From (1) and (2),

v3 DE = DC / v3

DE v3 * v3 = DC

3DE = DC

3DE = DE + EC

3DE – DE = EC

2DE = EC

DE = 50 / 2

DE = 25m

Height of the tower = DC

= DE + EC

= 25 + 50

= 75m

tan 45o = 20 / x

x = 20m = height of the building

tan 60o = h + 20 / x

v3 = h + 20 / 20

h = 20 (v3 - 1)

Height of the flag = 14.28m

x2 – [11x / 4] + [15 / 8] = 0

8x2 – 22x + 15 / 8 = 0

8x2 – 22x + 15 = 0

8x2 – 12x - 10x + 15 = 0

4x (2x – 3) – 5 (2x – 3) = 0

(2x – 3) (4x – 5) = 0

2x – 3 = 0 Or 4x – 5 = 0

x = 3 / 2 or x = 5 / 4

 Let n = a / [y + z] ; n = b / [z + x] ; n = c / [x + y]

y + z = a / n —- (1)

z + x = b / n —- (2)

x + y = c / n —- (3)

(2) – (1)

x – y = [b – a] / n

(a – b) = -n (x – y)

(3) – (2)

y – z = [c – b] / n

(b – c) = -n (y – z)

(1) – (3)

z – x = [a – c] / n

(c – a) = -n (z – x)

Now consider,

a (b – c) / y2 – z2

= a * (-n [y – z] / [y + z] [y – z]

= – na / y + z

= – n * n

= – n2

Similarly,

b (c – a) / z2 – x2

= b * (-n [z – x] / [z + x] [z – x]

= – nb / z + x

= – n * n

= – n2

c (a – b) / x2 – y2

= c * (-n [x – y] / [y + x] [x – y]

= – nc / x + y

= – n * n

= – n2

Therefore,

a (b – c) / y2 – z2 = b (c – a) / z2 – x2 = c (a – b) / x2 – y2

x = 4ab / a + b

To find [x + 2a] / [x – 2a] + [x + 2b] / [x – 2b]

x = 4ab / [a + b]

x = 2a x 2b / a + b

x / 2a = 2b / a + b

By componendo dividend,

(x + 2a) / (x – 2a) = (2b + a + b) / (2b – a – b)

Simplify RHS.

(x + 2a) / (x – 2a) = (a + 3b) / (b – a)

Similarly (x + 2b) / (x – 2b) = (3a + b) / (a – b)

(x + 2a) / (x – 2a) + (x + 2b) / (x – 2b)

= (a + 3b) / (b – a) + (3a + b) / (a – b)

= (a + 3b) / (b – a) – (3a + b) / (a – b)

= a + 3b – 3a – b / b – a

= 2b – 2a / b – a

= 2 (b – a) / b – a

= 2

 x + ay =b …. (i)

ax – by = c ….(ii)

Multiply equation (i) by a,

ax + a2y = ab …. (iii)

Subtracting equation (ii) from equation (iii),

a2y + by = ab – c

y = [ab – c] / [a2 + b]

Put the value of y in the equation (i),

x = b – ay

x = b – a {[ab – c] / [a2 + b]}

= [b2 + ac] / [a2 + b]

Let the numerator be = x

Let the denominator be = y

Therefore, the fraction is = x / y.

Given that,

(x – 5) / (y + 5) = 1 / 7

(7) (x – 5) = (1) (y + 5)

7x – 35 = y + 5

7x – y = 35 + 5

7x – y = 40 —- (1)

(x – 4) / (y) = 1 / 3

(3)(x – 4) = y

(3x – 12) = y

3x – y = 12 —- (2)

3 * (7x – y = 40)

7 * (3x – y = 12)

21x – 3y = 120 —- (3)

21x – 7y = 84 —-(4)

4y = 36

y = 36 / 4

y = 9

Put y = 9 in (1),

7x – y = 40

7x – 9 = 40

7x = 40 + 9

7x = 49

x = 49 / 7

x = 7

Thus the fraction is 7 / 9.

3x + 2y = 14 —- (1)

-x + 4y = 7 —- (2)

Multiply equation (2) by 3,

-3x + 12y = 21 —- (3)

3x + 2y = 14 —- (1)

-3x + 12y = 21 —- (3)

14y = 35

y = 35 / 14

y = 5 / 2

Put y = 5 / 2 in (1),

3x + 2 * (5 / 2) = 14

3x + 5 = 14

3x = 9

x = 3

There are 366 days in a leap year.

52 weeks + 2 days.

These 2 days can be:

(a) Monday, Tuesday

(b) Tuesday, Wednesday,

(c) Wednesday, Thursday

(d) Thursday, Friday

(e) Friday, Saturday

(f) Saturday, Sunday

(g) Sunday, Monday.

 Let A be the event that a leap year contains 53 Thursdays.

n (A) = {Wednesday, Thursday},{Thursday, Friday}}.

= 2

Required probability p (A) = n (A) / n (S)

= 2 / 7