SET - A

Any method mathematically correct should be given full credit of marks.

SECTION - AQuestion number 1 to 4 carry 1 marks each.

1.sec 35º

cosec 55º

=cosec (90 – 35)º

cosec 55º

=cosec 55ºcosec 55º

= 1

sec 35ºcosec 55º = 1 [1]

2. There are 26 letters in English alphabet. Let A be the event of choosing a letter preceding p.Number of favorable outcomes to A = 15.

P(A) =1526

The probability that the letter chosen precedes p is1526

. [1]

3. 5005

4. In ABC,DE || BC ... (given)

AD AE=DB EC ... (by basic proportionality

theorem)

MT EDUCARE LTD.CBSE X

Date :

SUBJECT : MATHEMATICS

QUEST - I (Semi Prelim I)MODEL ANSWER PAPER

SET A

Marks : 80

Time : 3 hrs.

50051001143131

11

57

13

B C

D E

A

1.5

cm

1 cm

3 cm

SET - A... 2 ...

1.5 1=3 EC

3 30EC = =

1.5 15 EC = 2 cm [1]

SECTION - BQuestion number 5 to 10 carry 2 marks each.

5. Ar (ABC) =12

[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

=12

[5(7 + 4) + 4(–4 – 2) + 7(2 – 7)]

=12

(55 –24 – 35)

=–42

= –2 As area is a measure, it cannot be negative, so we will take its numerical value. Area of ABC is 2 sq.units [1]

6. In ABC, ABC = 90o

tan 60o =ABBC

3 =AB15

AB = 15 m The height of the tower is 15 m. [1]

7. AB = 10

By distance formula, AB = 2 22 1 2 1( – ) + ( – )x x y y

10 = 2 2(11– 3) + ( +1)y 100 = (8)2 + (y + 1)2 … (Squaring both sides) 100 = 64 + y2 + 2y + 1 y2 + 2y + 65 – 100 = 0 y2 + 2y – 35 = 0

A

B C60º

15 m

SET - A... 3 ...

y2 + 7y – 5y – 35 = 0 y(y + 7) – 5(y + 7) = 0 (y + 7) (y – 5) = 0 y + 7 = 0 or y – 5 = 0 y = –7 or y = 5 The values of y are 7 and –5. [2]

8. x – y = 3 ... (i)

3x

+2y

= 6 ... (ii)

From (i), we get x = y + 3 … (iii)Substitute (iii) in (ii), we get

33

y +

2y

= 6

2(y + 3) + 3y = 36 2y + 6 + 3y = 36 5y = 30 y = 6

Substitute y = 6 in (iii), we get x = 6 + 3 = 9 x = 9 and y = 6 is the solution of given pair of linear equations. [2]

9. HCF of two numbers = 16 HCF × LCM = Product of two numbers

16 × LCM = 3072

LCM =3072

16= 192 [2]

10.2 0 2 0

2 0 2 0

sin 63 + sin 27cos 17 + cos 73 =

2 2

2 2 0

cos (90º – 63º) + sin 27ºsin (90º – 17º) + cos 73 ....

sin cos (90º – )and cos sin (90º – )

=2 0 2 0

2 0 2 0

cos 27 + sin 27sin 73 + cos 73

.... [sin2 + cos2 = 1]

=11

= 1

0 0

0 0sin² 63 + sin² 27cos² 17 + cos² 73 = 1 [2]

SET - A... 4 ...

11. In ABC,DE AC ... (given)

BE BD=EC DA ... (i)(by basic proportionality theorem)

In ABE,DF AE ... (given)

BF BD=FE DA ... (ii)(by basic proportionality theorem)

BF BE=FE EC ... from (i) and (ii) [2]

12. As the queen is drawn and kept aside, we have 4 cards left. So, total number of possible outcomes = 4 (a) Let E be the event that second card is an ace. Number of favorable outcomes to E = 1

P(E) =14

The probability that second card picked is ace is14 .

(b) Let F be the event that second card is an queen. Number of favorable outcomes to F = 0

P(F) =04 = 0

The probability that second card picked is queen is 0. [2]

SECTION - CQuestion numbers 13 to 22 carry 3 marks each.

13. Let the points R (a , b) and S (c , d) be the points of trisection of seg PQ.P (4, –1), Q (–2, –3)

R (a , b) divides PQ in the ratio 1 : 2 m1 = 1 , m2 = 2 By using section formula, we get

a = 1 2 2 1

1 2

++

m x m xm m b = 1 2 2 1

1 2

++

m y m ym m

P

(4, – 1)

R S Q

(c, d)

(a, b) (–2, –3)

SET - A... 5 ...

a =1 × (– 2) + 2 × 4

1 + 2 b =1 × (–3) + 2 × (–1)

1 + 2

=– 2 + 8

3=

–3–23

=63 b =

–53

a = 2

The coordinates of R =–

52,3

S (c, d) divides PQ in the ratio 2 : 1

c = 1 2 2 1

1 2

++

m x m xm m d = 1 2 2 1

1 2

++

m y m ym m

=2 × (– 2) + 1 × (4)

2 + 1 =2 × (–3) + 1 × ( –1)

2 + 1

=–4 + 4

3 =–6 – 1

3

=03 d =

–73

c = 0

The coordinates of S =

–70,3 [3]

14. Let the height of building (CD) = ‘h’ mHeight of tower (AB) = 50mIn BAC, BAC = 90º

tan 600 =ABAC

3 =50x

x =503

x =50 33 3 .... (Rationalising the denominator)

x =50 3

3.... (i)

A

B

x

D

C

h600300

50m

SET - A... 6 ...

In ACD, ACD = 90º

tan 300 =CDAC

13 =

hx

h =13 × x

h =13 ×

50 33

.... [from (i)]

h = 503

=2163

h = 16.67m

Therefore, height of the building is 16.67m [3]

OR

14. Let AB be the tree broken at C.Length of broken part = BC.Length of unbroken part= AC Height of tree = AC + BC

In right CAB,

tan 300 =ACAB

13 =

AC8

AC =83

In right CAB,

cos 300 =ABBC

32

=8

BC

3 BC = 16

BC =16

3

300

8m

BrokenPart

BA

C

B

SET - A... 7 ...

Height of the tree = AC + BC

=83 +

163 =

8 + 163 =

243

=243 ×

33

=24 3

3= 8 3 m

Height of the tree is 8 3 m [3]

15. Dividend = Divisor × Quotient + Remainder x³ – 3x² + x + 2 = g(x) × (x – 2) + (– 2x + 4)... g(x) × (x – 2) = x³ – 3x² + x + 2 + 2x – 4

... g(x) =3 23 3 2

( 2)x x x

x

x² – x + 13 22 3 3 2x x x x

x3 – 2x² – +

– x² + 3x – 2 – x² + 2x + –

x – 2x – 2

– + 0

g(x) = x² – x + 1 [3]

16. Let the digit in ten’s place be y, andthe digit in unit’s place be xOriginal number = 10y + xNumber obtained by interchanging the digits = 10x + yAccording to the first condition, According to the second condition,

x + y = 9 ...... (i) 9 (10y + x) = 2 (10x + y) 90y + 9x = 20x + 2y– 11x + 88y = 0 –x + 8y = 0 x – 8y = 0 ..... (ii)

SET - A... 8 ...

Subtracting eqn (ii) from eqn (i), we get Substituting y = 9 in eqn (i), we getx + y = 9 x + 1 = 9x – 8y = 0 x = 8– + –

9y = 9y = 1

Original number = 10y + x = 10 (1) + 8 = 10 + 8 = 18 Required two digit number is 18. [3]

OR

16. Let the length of a rectangle. be x unitsand the breadth of a rectangle be y units.Area of the rectangle = xy sq.unitsAccording to the first condition, According to the second condition,

(x – 5) (y + 3) = xy – 9 (x + 3) (y + 2) = xy + 67xy + 3x – 5y – 15= xy – 9 xy + 2x + 3y + 6= xy + 67

3x – 5y = 6 ......(i) 2x + 3y = 61 ......(ii)

Multiplying eqn (i) by 2 and eqn(ii) Substituting y = 9 in eqn (i)by 3 and subtracting them. 3x – 5(9)= 6 6x – 10y = 12 3x – 45= 6 6x + 9y = 183 3x = 51

– – – x= 17 – 19 y = – 171

y = 9

Length of the rectangle is 17 units, andBreadth of the rectangle is 9 units. [3]

17. Given : ABCD is a squareAQB is equilateral triangle on side AB.BRD is equilateral triangle on side BD.

Prove that : ar (AQB) =12 ar (BRD)

Proof :ABCD is a square ... (given)Let side of square be x units AB = BC = CD = AD = x unitsIn BADBAD = 90o ... (angle of a square)

Q

A

D

2

R

x

x

x

B

C

x

SET - A... 9 ...

BD2 = AB2 + AD2 ... (Phythagoras theorem) BD2 = x2 + x2

BD2 = 2x2

AQB ~ BRD ... (two equilateral triangles are always similar)

ar ΔAQBar ΔBRD =

2

2ABBD

...

ar ΔAQBar ΔBRD =

2

22xx

... ( AB = x , BD = 2.x )

ar ΔAQBar ΔBRD =

12

ar (AQB) =12 ar (BRD) [3]

18.

Proof :ABC ~ PQR .... (given)

ABPQ =

BCQR ...(i) .... (corresponding sides of similar

triangles)BC = 2 BD ...(ii) .... (seg AD is the median)QR = 2 QM ...(iii) .... (seg PM is the median)ABPQ =

2BD2QM ...(iv) ... from (i), (ii), (iii)

B = Q ...(v) .... (corresponding angles of similar triangles)

In ABD & PQM,

ABPQ =

BDQM .... from (iv)

B = Q .... from (v) ABD ~ PQM .... (by SAS similarity criterion)

ABPQ =

ADPM .... (corresponding sides of similar triangles) [3]

RMQ

PA

B D C

SET - A... 10 ...

19. L.H.S. =cosA–sinA+1cosA+sinA–1

Dividing Numerator & Denominator by sin A

=cotA – 1 + cosecAcotA + 1 – cosecA

=2 2(cotA + cosec A) – (cosec A – cot A)

(1 + cotA–cosecA) .. ( cosec² A – cot² A = 1)

=(cotA + cosec A) – (cosec A + cot A) (cosec A – cot A)

(1+cotA – cosecA)

=(cotA+cosec A) (1-cosec A+cot A)

(1+cotA-cosecA)

= cot A + cosec A= R.H.S.

L.H.S. = R.H.S.

cosA – sinA+1cosA + sinA–1 = cosec A + cot A using the identity

cosec² A = 1 + cot² A. [3]

OR

19. L.H.S= (sin A + cosec A)2 + (cos A + sec A)2

= sin2A + 2sinA cosecA + cosec2A + cos2A + 2cosA secA + sec2A= (sin2A + cos2A) + (cosec2A )+ (sec2A) + 2sinA cosecA + 2 cosA secA

= 1+ (1+ cot2A) + (1 + tan2A) + 2 sinA1 1+2cos A

sin A cos A

.... [ sin² A + cos² A = 1, cosec² A = 1 + cot² A, sec² A = 1 + tan² A]= 1 + 1 + cot² A + 1 + tan² A + 2 + 2= 7 + tan2A + cot2A= R.H.S.

L.H.S. = R.H.S. (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A [3]

SET - A... 11 ...

20. A die is thrown twice Total no. of possible outcomes

= { (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) } = 36

(i) Let A be the event that 5 will not come up either time No. of outcomes favourable to A

= { (1, 1) (1, 2) (1, 3) (1, 4) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 6) } = 25

P(A) =2536

(ii) Let B be the event that 5 will come up at least once No. of outcomes favourable to B

= { (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)(5, 1), (5, 2), (5, 3), (5, 4), (5, 6) } = 11

P(B) =1136 [3]

OR

20. Total no. of possible outcomes = 90(i) Let A be the event that the disc drawn bears a two digit number.

= {10, 11, 12,.........., 90} = 81 No. of outcomes favourable to A = 81

P(A) =81

90 =

910

(ii) Let B be the event that the disc bears a perfect square number= {1, 4, 9, 16, 25, 36, 49, 64, 81} = 9

No. of outcomes favourable to B = 9

P(B) =990 =

110

SET - A... 12 ...

(iii) Let C be the event that the disc drawn bears a number divisible by 5= { 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}= 18

No. of outcomes favourable to C = 18

P(B) =18

90 =

15

[3]

21. The given system of equations is3x – 4y + 7 = 0kx + 3y – 5 = 0This is of the form a1x + b1y + c1 = 0

a2x + b2y + c2 = 0Where, a1 = 3, b1 = –4, c1 = 7 and a2 = k, b2 = 3, c2 = –5For no solution, we must have

1

2

aa = 1

2

bb 1

2

cc

We have, 1

2

bb

=–43

and 1

2

cc

=–75

Clearly, 1

2

bb

1

2

cc

So, the given system will have no solution.

If1

2

aa = 1

2

bb

3k =

–43

k =–94

Clearly, for this value of k, we have 1

2

aa

= 1

2

bb 1

2

cc

Hence, the given system of equations has no solution, when k =–94 [3]

22. Divisor × Quotient + Remainder= (–x2 + x – 1) (x – 2) + 3= –x3 + x2 – x + 2x2 – 2x + 2 + 3= –x3 + 3x2 – 3x + 5= Dividend

The division alrorithm is verified.[3]

x – 22 3 21 – 3 – 3 5x x x x x

–x3 + x² – x + – +

2x² – 2x + 5 2x² – 2x + 2 – + –

3

SET - A... 13 ...

SECTION - DQuestion numbers 23 to 30 carry 4 marks each.

23. Given : ABC is an equilateral triangle.

D is a point on BC such that BD =1

BC3

Prove that : 9AD2 = 7AB2

Construction : Draw AP BCProof :ABC is an equilateral triangle ... (given)

AP BC ... (construction)B = 600 ... (angle of an equilateral

triangle)In right APB,

sin 600 =APAB

32

=APAB

AP =32

AB ... (I)

Also,BP = CP

(altitude of an equilateral triangle drawn to the base bisects it)

P is midpoint of BC BP =12 BC ... (II)

BD =13 BC ... (III) .... (given)

PD = BP – BD

PD =1 BC2

–1 BC3 .... from (II) & (III)

PD =1 BC6

PD =1 AB6 ...(IV).... ( AB = BC = AC)

In right APD,AD² = AP² + PD² ... (by Pythagoras theorem)

=2

3 AB2

+21 AB

6

... [from (I) and (IV)]

A

B CPD600

SET - A... 14 ...

=3 AB²4

+136 AB²

=27 AB²36 +

136 AB²

=27 + 1 AB²

36

=28 AB²36

AD² =7 AB²9

9AD2 = 7AB2 [4]

24. Let the present age of Aftab be x years,and the present age of his daughter be y yearsSeven years ago age of Aftab was (x – 7) yearsand the age of his daughter was (y – 7) yearsAccording to the first condition,

x – 7 = 7 (y – 7) x – 7 = 7y – 49x – 7y – 7 + 49= 0 x – 7y + 42 = 0 .... (i)After three years age of Aftab will be (x + 3) years,and the age of his daughter will be (y + 3) years According to the second condition,

(x + 3) = 3 (y + 3) x + 3 = 3y + 9 x – 3y + 3 – 9 = 0 x – 3y – 6 = 0 .... (ii)So, Algebraically the two situations are expressed as

x – 7y + 42 = 0 .... (i)x – 3y – 6 = 0 .... (ii)

Now to represent graphically we find two solutions of each equation.x – 7y + 42 = 0 x – 3y – 6 = 0

Now we plot the points R (0, 6), S (7, 7), P (0, –2) and Q (6 , 0)

x 0 7

y 6 7

x 0 6

y – 2 0

SET - A... 15...

[4]

OR

24. Let the speed of train be x km/hr andthe speed of bus be y km/hrAccording to the first condition,When Roohi travels 60 km by train and travels 240 km by bus, shetakes 4 hrs.

60x +

240y = 4 ........

DistanceTime =Speed

According to the second condition,When Roohi travels 100 km by train and 200km by bus,

ScaleOn X axis 1 cm = 1unit

YOn axis 1 cm = 1 unit

-7

-6

-5

-4

-3

-1

-1-2-3-4-5-6 8764321

-2

5

1

2

3

4

5

6

7

8

x y-7 + 42

xy– 3 – 6 = 0

P (0, -2)

R (0, 6)

S (7, 7)

Q (6, 0) XX'

Y'

Y

SET - A... 16 ...

she takes 4hr + 10 mins. =10460 = 6

25 hrs. [ 1hr = 60 mins.]

100

x +200y

= 625

........DistanceTime =

Speed

Substituting1x = u &

1y = v

60 u + 240 v = 4 ........ (i)

100 u + 200 v = 625

........ (ii)

Multiplying (i) by 10 & (ii) by 6 600 u + 2400 v = 40 ... (iii) 600 u + 1200 v = 25 ... (iv)

Subtracting eqn (iv) from eqn (iii) Substituting v = 801

in eqn (i)

600 u + 2400 v= 40 60 u + 240 × 801

= 4

600 u + 1200 v = 25 60 u + 3 = 4– – –

1200 v = 15 60u = 1

v = 120015

u =160

v =180

Resubstituting for p and q.

1x = u

1y = v

1x = 60

1

1y = 80

1

x = 60 y = 80

Speed of the train is 60km/hr andSpeed of the bus is 80km/hr [4]

25. Let the height of tower (AB) = ‘h’ mLet the distance of the foot of tower to the car at C = (AC) = ‘y’ mLet the distance covered by the carin 6 seconds (CD) = ‘x’ mIn BAC, BAC = 90º

SET - A... 17 ...

tan 600 =ABAC

3 =hy

h = 3y .... (i)In BAD, BAD = 90º

tan 300 =ABAD

13 =

hx y

x + y = h 3 .... (i)

x + y = 3 y × 3 .... [from (i)] x + y = 3y x = 2y .... (ii)Distance covered by car in 6 seconds = x m

Speed of the car =Distance

Time

= 6x

=26y

.... [from (ii)]

= m/sec3y

Time taken by car to reach the foot of tower from point C,

Time =Dis tance

Speed

=3

yy

= y ×3y

= 3 sec.Time taken by the car to reach the foot of tower from thispoint is 3 sec. [4]

OR

25. Let the height of tower (AB) = ‘h’mDistance of point C from the base of tower = 4mDistance of point D from the base of tower = 9m

B

DCA y x

600

300

600 300

h

E

SET - A

Let ACB = So, ADB = (90 – )In 1st case :In right BAC,

tan =ABAC

tan =4h

..... (i)

In 2nd case :In right BAD,

tan (90 – ) =ABAD

tan (90 – ) = 9h

cot = 9h

..... [tan (90 – ) = cot ]

1

tan = 9h

1h4

= 9h

4h = 9

h

36 = h²i .e. h = 6m

Height of the tower is 6m [4]

26. Given : A ABC, in which DE BC suchthat DE intersects AB and

AC at D and E respectively.

To Prove :AD AE=DB EC

Construction : Join BE and CD and draw EF AB, DG AC.

Proof : We have :

ar (ADE) =12 × AD × EF

... 18 ...

B

h

A

CD90 – 5m 4m

9m

A

ED

B C

GF

SET - A

andar (DBE)=12 × DB × EF

ar ADEar DBE =

1 × AD × EF21 × DB × EF2

=ADDB

.... (i)

Again,ar (ADE) =12 × AE × DG

and ar (DCE) =12 × EC × DG ...

ar ADEar DCE =

1 × AE × DG21 × EC × DG2

=AEEC .... (ii)

Now, DBE and DCE are on the same base DE and between thesame parallels DE and BC. ar (DBE) = ar (DCE)Therefore, equation (ii) becomes,

ar ADEar DBE =

AEEC .... (iii)

From (i) and (iii), we get :AD AE =DB EC [4]

27. Let x be any positive integer and b = 3Applying Euclid’s Division Algorithm x = 3q + r where 0 < r < 3 The possible remainders are 0, 1, 2 x = 3q or 3q + 1 or 3q + 2(i) If x = 3q x3 = (3q)³ = 27q³ = 9(3q³) = 9m for some integer m,

where m = 3q³

(ii) If x = 3q + 1 x³ = (3q + 1)³ = (3q)3 + 3(3q)²(1) + 3(3q) (1)² + (1)³[ since (a +b) ³ = a³ + 3a²b + 3ab² + b³]

= 27q3 + 27q2 + 9q + 1= 9q(3q² + 3q + 1) + 1= 9m + 1 for some integer m,

where m = q (3q² + 3q + 1)

... 19 ...

SET - A

(iii) If x = 3q + 2 x3 = (3q + 2)3

= (3q)3 + 3(3q)² (2) + 3(3q) (2)² + (2)3

[ since (a + b)³ = a³ + 3a²b + 3ab² + b³]= 27q3 + 54q2 + 36q + 8= 9q(3q² + 6q + 4) + 8= 9m + 8 for some integer m,

where m = q (3q² + 6q + 4)Hence, the cube of any positive integer is either of the form9m, 9m + 1 or 9m + 8 [4]

28. A (– 3, 5) , B (3, 1), C (0, 3), D (– 1, – 4)

AB = 2 23 ( 3) (1 5) CD = 2 2(–1–0) + (–4–3)

= 2 26 + (– 4) = 2 2(–1) + (–7)

= 36 + 16 = 1 + 49

= 52 = 50

= 2 13 units = 5 2 units

BC = 2 2(0 – 3) + (3 – 1) AD = 2 2(–1 + 3) + (– 4 – 5)

= 2 2(–3) + (2) = 2 2(2) + (– 9)

= 9 + 4 = 814

= 13 units = 85 units

Diag AC = 2 2(0 + 3) + (3 – 5) Diag BD = 2 2(–1 – 3) + (– 4 – 1)

= 2 2(3) + (– 2) = 2 2(– 4) + (– 5)

= 9 + 4 = 16 + 25

= 13 units = 41 units

As AC + BC = AB, The given points do not form any quadrilateral [4]

29. Substituting1

(3 )x y = p &1

(3 )x y = q

p + q =43

2p

–2q

= 81 2

p q = 8

1

4p + 4q = 3 8p – 8q = – 2

... 20 ...

SET - A

p =3 4

4– q

...... (i) 4p – 4q = – 1 ..... (ii)

Substituting eqn (i) in eqn (ii) Substituting q = 21

in eqn (i)

3 44

4– q

– 4q = – 1 4p – 412

= – 1

12 16

4– q

– 4q = – 1 4p – 2 = – 1

12 – 16q – 16q = – 4 4p = 1

– 32q = – 16 p = 14

q =12

Resubstituting for p and q.

p =1

3 x y q =1

3 x y

41

=1

3 x y 21

=1

3 x y 3x + y = 4 ...(iii) 3x – y = 2 ...(iv)

Adding eqn (iii) and eqn (iv) Substituting x = 1 in eqn (i) 3x + y = 4 3 (1) + y = 4 3x – y = 2 y = 1 6x = 6 x = 1 Solution is x = 1 , y = 1 [4]

OR

29. 7y

–2x = 5 ;

8y +

7x = 15

Substituting1y

= p &1x = q

7p – 2q = 5 8p + 7q = 15 ... (ii)

p =5 2

7 q

... (i)

Substituting eqn (i) in eqn (ii) Substituting q = 1 in eqn (i)

85 2

7q

+ 7q= 15 p =5 + 2(1)

7

... 21 ...

SET - A

40 + 16 + 49

7q q

= 15 p = 1

40 + 65q = 105 65q = 65 q = 1Resubstituting for p and q.

p =1y q =

1x

1 =1y 1 =

1x

y = 1 x = 1Solution is x = 1, y = 1 [4]

30. Let the distance he walked towards the building = ‘x’ mHeight of tower (AB) = 30mHeight of boy (CD) = 1.5mbut, CD = EF = AGAG = 1.5mBG = AB – AGBG = 30 – 1.5BG = 28.5mIn BGF, BGF = 90º

tan 600 =BGGF

3 =28.5

y

y =28.5

3

y =28.5 3

3 3 (Rationalising the denominator)

y = 28.5 × 33

y = 9.5 3 ..... (i)In BGD, BGD = 90º

tan 300 =BGGD

B

G

A E C

DF 300600

28.5

1.5m

1.5m

1.5m

x

xy

... 22 ...

SET - A

13 =

28.5y x

y + x = 28.5 3

x = 28.5 3 – y

x = 28.5 3 – 9.5 3 [From (i)]

x = 3 (28.5 – 9.5)

x = 19 3 x = 19 × 1.73 x = 32.87 The distance walked by the boy is 32.87 m. [4]

... 23 ...