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
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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]
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