Section-1
1). Evaluate cosec 39deg . sec 51deg - tan 51deg . cot 39deg
sol) cosec ( 90 - 51) . sec 51 - tan 51 . cot(90-51)
**********************************************
We know cosec(90-0) = sec 0 and cot(90 - 0) = tan 0
**********************************************
=> sec 51 . sec 51 - tan 51 . tan 51
=> sec^2(51) - tan^2(51)
***************************
we know sec^20 - tan^2 0 = 1
***************************
=> 1
**************************************************************************
2) Write the similarity criterion by which the given pair of triangles are similar.
sol) Finding the ratio of sides
AB = 2 = 1
QR 4 2
AC = 3 = 1
QP 6 2
BC = 2.5 = 1
RP 5 2
Hence, AB = AC = BC = 1
QR QP RP 2
SSS Similarity Theorem:- if all 3 pairs of corresponding sides of two triangles are proportional, then the two triangles are similar.
.^. Using similarly SSS criterion
So, △ABC ~ △QRP
*************************************************
3). From English alphabet if a letter is chosen at random, the find the probability that the letter is consonant.
sol) Total English alphabets = 26
Number of consonants = 21
.^. the probability that the letter is consonant(P) = 21
26
*****************************************************
4). In a right triangle ABC, right angled at "C" in which AB= 13 cm, BC = 5 cm, determine the value of cos^2 B + Sin^2 A
sol) Using Pythagoras theorem
AB^2 = AC^2 + BC^2
(13)^2 = AC^2 + (5)^2
169 = AC^2 + 25
169-25 = AC^2
144 = AC^2
√144 = AC {12*12 = 144}
.^. AC = 12 cm
Now,
cos^2 B + Sin^2 A
we know cos0 = adj
hyp
cos^2 B = Adj = BC = (5/13)^2
Hyp AB
sin^2 A = Opp = AC = (12/13)^2
Hyp AB
cos^2 B + Sin^2 A ;-
(5/13)^2 + (12/13)^2
(25/169) + (144/169)
25 + 144
169
=>169
169
cos^2 B + Sin^2 A = 1
*************************************************
(5) A point "P" is 25 cm from the center "O" of the circle. The length of the tangent drawn from "P" to the circle is 24 cm. Find the radius of the circle.
sol)
Using Pythagoras Theorem:-
OP^2 = OQ^2 + QP^2
=> (25)^2 = (x)^2 + (24)^2
=> 625 - 576 = x^2
=> 49 = x^2
=> x= √49 {7 * 7 = 49}
=> x or radius(r) = 7
*****************************************************
6) Find the Median of first seven composite numbers
sol) Composite Numbers:- A whole number that can be made by multiplying other whole number.
excluding prime numbers which can only be formed by multiplying with"1"
example:- 2*2 = 4, 2*3=6 but not "7" as its prime
First seven composite Numbers :-
2*2 = 4
2*3=6
2*4=8
3*3 =9
2*5 = 10
2*6 = 12
2*7 =14
4, 6, 8, 9, 10, 12, 14
.^. the median of 1st seven composite number is "9"
***************************************************
7) In a hemispherical bowl of 2.1 cm radius ice cream is there. Find the volume of the bowl.
sol) Given :- Radius of bowl = 2.1 cm
We have,
Volume of hemisphere = (2/3) *𝛱 * (r)^3
=> 2 * 22 * (2.1)^3
3 7
=> (0.6) * (3.14 ) * 9.26
=> (1.88) * ( 9.26)
Hence, the volume of the spherical bowel =17. 40 cm^3
************************************************
Section-2
8) Write the mode formula for grouped data and explain the terms in it.
sol )
Modal class = Interval with highest frequency
L = Lower limit of modal class
h= class-interval
F1 = Frequency of modal class
Fo= Frequency of class before modal class
F2 = Frequency of class after modal class
*************************************************
9) In the given figure, TA and TB are tangents to the circle with center "O" . If ∠ATB = 80 deg, then find the measure of ∠ABT
sol) Given :- TA and TB are tangents
We Know tangents drawn to the circle from same point is always equal
so, tangent (1) TA = TB Tangent(2)
Given :- ∠ATB = 80 -------(1)
we can see from the figure in triangle TAB
we know, sum of angles in a triangle = 180
∠ATB + ∠TBA +∠BAT = 180
from (1) , ∠ATB = 80
80 + ∠TBA +∠BAT = 180
Since TA = TB opposite angles of equal sides are equal
80 + 2ATB = 180
2ATB = 180 - 80
ATB = 100 / 2
ATB = 50 deg
***********************************************
10) A bag contains balls numbered 1-50, a ball is drawn at random from the bag, the probability that it bears a two digit numbered multiple of 7
sol) Two digit numbers that are between 1-50 but divisble by "7" are as follows:
14, 21, 28, 35, 42, 49 = total 6
we have,
Probability = number of events
total possible outcomes
Total possible outcome = 50
No.of events = 6
.^. probability = 6 simplify further= 3
50 25
*****************************************************
11) From the top of the building the angle of elevation of the top of the cell tower is 60deg and the angle of depression to its foot is 45deg, if the distance of the building from the tower is 30 meters. draw the suitable diagram to the given data.
sol)
condition (1) :- From the top of building, the angle of elevation of the top of a cell tower is 60 deg = ∠AEB
condition(2) :- The angle of depression to its foot is
45 deg =∠ECB
condition(3) :- Distance of the building from the tower is
EB=DC = 30m
In ⧍AEB
Tan 𝜽 = opp
adj
Tan 60 = AB
EB
√3 = AB
30
30√3 = AB
(*) In ⧍ EDC
Tan 𝜽 = opp
adj
Tan 45 = ED
DC
1 = ED
30
30 = ED
SO, ED = BC = 30 m
Height of the tower = AB + BC
=> 30 √3 + 30
=> (30 * 1.73) + 30
=> (51.9) + 30
.^. 81.9m is the height of the tower
***************************************************
12) Find the value of
tan^2 60 + cot^2 30
sin^2 30 + cos^2 60
sol) (√3)^2 + (√3)^2
(1/2)^2 + (1/2)^2
=> 3 + 3
(1/4) + (1/4)
=> 6
1 + 1
4
=> 6
(2/4)
=> 6 * 4
2
=>24
2
=> 12
*****************************************************
13). A right circular cylinder has radius 3.5 cm and height 14cm. Find curved surface area.
sol) We have :- C.S.A = 2𝛑 r h
=> 2 * (22/7) * (3.5) * (14)
=> 2 * 3.14 * 49
=> 153.86 * 2
.^. C.S.A = 307.72 cm^2
*************************************************
Section-3
sol) Steps to draw ⧍ PQR
1) Draw base QR of side 6 cm
Q-------------6cm--------------R
2) Draw ∠Q = 70 deg
3) Taking "Q" as center, 4cm as radius, we draw an arc .
let the point where arc intersects the ray be point "P"
4) Join PR
.^. ⧍PQR is the required triangle
Now, we need to make a triangle which is 3/4times its size
.^. scale factor = (3/4) < 1
(*) Steps of construction
1) Draw any ray QX making an acute angle with QR on the side opposite to the vertex "P"
2) Mark 4( the greater of 3 and 4 in (3/4)) points
Q1,Q2,Q3,Q4 on QX so that QQ1=Q1Q2=Q2Q3=Q3Q4
3) Join Q4R and draw a line through Q3( the 3rd point, 3 being smaller of 3 and 4 in 3/4) parallel to Q4R, to intersect QR at R'.
4) Draw a line through R' parallel to the line PR to intersect QP at P'
Thus, ⧍ P'QR' is the required triangle
***********************************************
14(b) Draw less than Ogive for the following frequency distribution. Find the median from obtained curve.
sol)
Less than type Ogive
obtaining median weight from the graph:-
Since Σ = 2+5+12+31+39+10+4 = 103
So, n = 103
(n/2) = 103/2 = 51.5
we draw a line parallel to x-axis where number of students = 51.5
we can see from the ogive it intersecting x-axis at "100"
.^. Median = 51.5
Verifying using median formula
L = Lower limit of median class = 90
class interval(h) = 70-60= 10
cumulative frequency of the class before median class(cf) = 20
Frequency of median class(f) = 31
90 + (51.5-20/31) * 10
=100.16 verified/-
****************************************************
15 (a) Show that cos𝜽 + 1 -sin𝜽 =2 sec𝜽
1-sin𝜽 cos𝜽
sol) cross multiply we get,
cos^2 𝜽 + (1-sin)^2 𝜽
(1-sin𝜽)(cos𝜽)
***********************************
(1-sin)^2 is in (a-b)^2 form expand we get
************************************
cos^2 𝜽 + (1 + sin^2𝜽 -2*1*sin𝜽)
(1- sin𝜽) cos𝜽
**********************
WE KNOW,
sin^2 𝜽 + cos^2 𝜽 = 1
***********************
(Sin^2 𝜽 + cos^2 𝜽) + 1 - 2sin𝜽
(1-sin𝜽) cos𝜽
1+ 1- 2sin𝜽)
(1-sin𝜽) cos𝜽
2-2 sin𝜽
(1-sin𝜽) cos𝜽
taking"2" common from numerator
2(1-sin𝜽)
(1-sin𝜽) cos𝜽
2
cos𝜽
2 * 1
cos𝜽
From Trigonometry identities we know ( 1 /cos𝜽 = sec𝜽)
.^. 2 sec𝜽
**************************************************
15(b) In a right angled triangle the hypotenuse is 10cm more than the shorter side. If third side is 6cm less than hypotenuse, find the sides of right angled triangle
sol)
condition 1) :- Hyp is 10cm more than shorter side
Let "x" be the shorter side
.^. Hyp = 10+x
cond 2) :- 3rd side 6cm less than Hyp
Hyp - 6cm
10+x-6 = x + 4 cm
By using Pythagoras theorem
AC^2 = AB^2 + BC^2
( x+10)^2 = ( x)^2 + ( x+4)^2
**********************************
(x+10)^2 and (x+4)^2 is in (a+b)^2 form
***************************************************
x^2 + 100 +20x = x^2 + ( x^2 +16+8x)
100 + 20x = x^2 + 16 + 8x
84 + 20x = x^2 + 8x
x^2 - 12x - 84 = 0
we have
x = -b + √(b^2 - 4ac)
2a
x = -(-12) + √(-12)^2 - 4 * 1 * (-84)
2 * 1
x = 12 + √144 + 336
2
x = 12 + √480
2
x = 12 + 21.90
2
x = (12 + 21.90)
2
=> (33.90)/2 = 16.95
or
x = 12 - 21.90
2
=> -4.95
As sides cannot b negative we take x= 19.95
now
Shorter side (x) = 16.95 cm
Third side (x + 4) = 20.95 cm
Hyp (x + 10) = 26.95 cm
***********************************************
16) Find the mean age of 100 residents of a colony from the following data
sol) Assumed Mean(A) = 35
𝝨 fi = 10 + 15 + 25 + 25+10+10+5 = 100
𝝨 fi*Ui = -30-30-25+0+10+20+15 = -40
Mean:-
Assumed meand(A) = 35
Class interval(h) = (20-10) = 10
𝚺fi = 100
𝚺fi * Ui = -40
X = a + h [ 𝞢 fi ui]
𝞢 fi
=> 35 + 10 [ -40 ]
100
=> 35 +(-4)
=35-4
=31 is the mean age of 100 residents
*********************************************
16(b) A toy is made with seven equal cubes of side √7 cm. six cubes are joined to six faces of a seventh cube find the total surface of the toy.
sol) Given :- No.of cube = 7
Side of the cube = s = √7 cm
Six cubes are joined to 6 faces of a 7th cube
if one cube attached with other cube one of its face get hidden and only five faces of cube are exposed and only surface area of six cube is effective
we have
Surface area of one side cube = S^2 = (√7)^2 = 7 cm^2
.^. , surface area of five side of a cube = 5 * 7 = 35 cm^2
.^. surface area of six cube = 6 * 35 = 210 cm^2
Total surface area of the toy = 210 cm^2
***************************************************
17(a) If two dice are thrown at the same time, find the probability of getting sum of the dots on top is prime.
sol) Prime Number:- which is divisible by 1 and itself
2,3,5,7,11,13, 15........
(1,1) ,( 1,2),(1,4), (1,6) = 4
(2,1),(2,3),(2,5) =3
(3,2), (3,4) =2
(4,1), (4,3) =2
(5,2), (5,6) =2
(6,1),(6,5) =2
----------------------
Total 15 outcomes
.^. probability for getting sum as prime number
=> No.of favourable event
Total No.of outcomes
=> 15
36
=> 5 is the required probability
12
***************************************************
17 (b) The angle of elevation of the top of a hill at the foot of a tower is 60 degree and the angle of elevation of the top of the tower from the foot of the hill is 30 deg. If the tower is 50m high, what is the height of the hill.
sol)
(!) In ⧍ ABC
AB = Tan 60
BC
50 = 1
BC √3
BC = 50 * √3 ---------------------(1)
(!!) In ⧍ DBC
DC = Tan 30 deg
BC
DC = √3
50√3 -------------->(From (1))
DC = 50√3 * √3
DC = 50 * 3
DC = 150 m is the height of the hill
1). Evaluate cosec 39deg . sec 51deg - tan 51deg . cot 39deg
sol) cosec ( 90 - 51) . sec 51 - tan 51 . cot(90-51)
**********************************************
We know cosec(90-
=> sec^2(51) - tan^2(51)
***************************
we know sec^2
***************************
=> 1
**************************************************************************
2) Write the similarity criterion by which the given pair of triangles are similar.
sol) Finding the ratio of sides
AB = 2 = 1
QR 4 2
AC = 3 = 1
QP 6 2
BC = 2.5 = 1
RP 5 2
Hence, AB = AC = BC = 1
QR QP RP 2
SSS Similarity Theorem:- if all 3 pairs of corresponding sides of two triangles are proportional, then the two triangles are similar.
.^. Using similarly SSS criterion
So, △ABC ~ △QRP
*************************************************
3). From English alphabet if a letter is chosen at random, the find the probability that the letter is consonant.
sol) Total English alphabets = 26
Number of consonants = 21
.^. the probability that the letter is consonant(P) = 21
26
*****************************************************
4). In a right triangle ABC, right angled at "C" in which AB= 13 cm, BC = 5 cm, determine the value of cos^2 B + Sin^2 A
sol) Using Pythagoras theorem
AB^2 = AC^2 + BC^2
(13)^2 = AC^2 + (5)^2
169 = AC^2 + 25
169-25 = AC^2
144 = AC^2
√144 = AC {12*12 = 144}
.^. AC = 12 cm
Now,
cos^2 B + Sin^2 A
we know cos
hyp
cos^2 B = Adj = BC = (5/13)^2
Hyp AB
sin^2 A = Opp = AC = (12/13)^2
Hyp AB
cos^2 B + Sin^2 A ;-
(5/13)^2 + (12/13)^2
(25/169) + (144/169)
25 + 144
169
=>
cos^2 B + Sin^2 A = 1
*************************************************
(5) A point "P" is 25 cm from the center "O" of the circle. The length of the tangent drawn from "P" to the circle is 24 cm. Find the radius of the circle.
sol)
Using Pythagoras Theorem:-
OP^2 = OQ^2 + QP^2
=> (25)^2 = (x)^2 + (24)^2
=> 625 - 576 = x^2
=> 49 = x^2
=> x= √49 {7 * 7 = 49}
=> x or radius(r) = 7
*****************************************************
6) Find the Median of first seven composite numbers
sol) Composite Numbers:- A whole number that can be made by multiplying other whole number.
excluding prime numbers which can only be formed by multiplying with"1"
example:- 2*2 = 4, 2*3=6 but not "7" as its prime
First seven composite Numbers :-
2*2 = 4
2*3=6
2*4=8
3*3 =9
2*5 = 10
2*6 = 12
2*7 =14
4, 6, 8, 9, 10, 12, 14
.^. the median of 1st seven composite number is "9"
***************************************************
7) In a hemispherical bowl of 2.1 cm radius ice cream is there. Find the volume of the bowl.
sol) Given :- Radius of bowl = 2.1 cm
We have,
Volume of hemisphere = (2/3) *𝛱 * (r)^3
=> 2 * 22 * (2.1)^3
3 7
=> (0.6) * (3.14 ) * 9.26
=> (1.88) * ( 9.26)
Hence, the volume of the spherical bowel =17. 40 cm^3
************************************************
Section-2
8) Write the mode formula for grouped data and explain the terms in it.
sol )
Modal class = Interval with highest frequency
L = Lower limit of modal class
h= class-interval
F1 = Frequency of modal class
Fo= Frequency of class before modal class
F2 = Frequency of class after modal class
*************************************************
9) In the given figure, TA and TB are tangents to the circle with center "O" . If ∠ATB = 80 deg, then find the measure of ∠ABT
sol) Given :- TA and TB are tangents
We Know tangents drawn to the circle from same point is always equal
so, tangent (1) TA = TB Tangent(2)
Given :- ∠ATB = 80 -------(1)
we can see from the figure in triangle TAB
we know, sum of angles in a triangle = 180
∠ATB + ∠TBA +∠BAT = 180
from (1) , ∠ATB = 80
80 + ∠TBA +∠BAT = 180
Since TA = TB opposite angles of equal sides are equal
80 + 2ATB = 180
2ATB = 180 - 80
ATB = 100 / 2
ATB = 50 deg
***********************************************
10) A bag contains balls numbered 1-50, a ball is drawn at random from the bag, the probability that it bears a two digit numbered multiple of 7
sol) Two digit numbers that are between 1-50 but divisble by "7" are as follows:
14, 21, 28, 35, 42, 49 = total 6
we have,
Probability = number of events
total possible outcomes
Total possible outcome = 50
No.of events = 6
.^. probability = 6 simplify further= 3
50 25
*****************************************************
11) From the top of the building the angle of elevation of the top of the cell tower is 60deg and the angle of depression to its foot is 45deg, if the distance of the building from the tower is 30 meters. draw the suitable diagram to the given data.
sol)
condition (1) :- From the top of building, the angle of elevation of the top of a cell tower is 60 deg = ∠AEB
condition(2) :- The angle of depression to its foot is
45 deg =∠ECB
condition(3) :- Distance of the building from the tower is
EB=DC = 30m
In ⧍AEB
Tan 𝜽 = opp
adj
Tan 60 = AB
EB
√3 = AB
30
30√3 = AB
(*) In ⧍ EDC
Tan 𝜽 = opp
adj
Tan 45 = ED
DC
1 = ED
30
30 = ED
SO, ED = BC = 30 m
Height of the tower = AB + BC
=> 30 √3 + 30
=> (30 * 1.73) + 30
=> (51.9) + 30
.^. 81.9m is the height of the tower
***************************************************
12) Find the value of
tan^2 60 + cot^2 30
sin^2 30 + cos^2 60
sol) (√3)^2 + (√3)^2
(1/2)^2 + (1/2)^2
=> 3 + 3
(1/4) + (1/4)
=> 6
1 + 1
4
=> 6
(2/4)
=> 6 * 4
2
=>
2
=> 12
*****************************************************
13). A right circular cylinder has radius 3.5 cm and height 14cm. Find curved surface area.
sol) We have :- C.S.A = 2𝛑 r h
=> 2 * (22/7) * (3.5) * (14)
=> 2 * 3.14 * 49
=> 153.86 * 2
.^. C.S.A = 307.72 cm^2
*************************************************
Section-3
sol) Steps to draw ⧍ PQR
1) Draw base QR of side 6 cm
Q-------------6cm--------------R
2) Draw ∠Q = 70 deg
3) Taking "Q" as center, 4cm as radius, we draw an arc .
let the point where arc intersects the ray be point "P"
4) Join PR
.^. ⧍PQR is the required triangle
Now, we need to make a triangle which is 3/4times its size
.^. scale factor = (3/4) < 1
(*) Steps of construction
1) Draw any ray QX making an acute angle with QR on the side opposite to the vertex "P"
2) Mark 4( the greater of 3 and 4 in (3/4)) points
Q1,Q2,Q3,Q4 on QX so that QQ1=Q1Q2=Q2Q3=Q3Q4
3) Join Q4R and draw a line through Q3( the 3rd point, 3 being smaller of 3 and 4 in 3/4) parallel to Q4R, to intersect QR at R'.
4) Draw a line through R' parallel to the line PR to intersect QP at P'
Thus, ⧍ P'QR' is the required triangle
***********************************************
14(b) Draw less than Ogive for the following frequency distribution. Find the median from obtained curve.
sol)
Less than type Ogive
obtaining median weight from the graph:-
Since Σ = 2+5+12+31+39+10+4 = 103
So, n = 103
(n/2) = 103/2 = 51.5
we draw a line parallel to x-axis where number of students = 51.5
we can see from the ogive it intersecting x-axis at "100"
.^. Median = 51.5
Verifying using median formula
class interval(h) = 70-60= 10
cumulative frequency of the class before median class(cf) = 20
Frequency of median class(f) = 31
90 + (51.5-20/31) * 10
=100.16 verified/-
****************************************************
15 (a) Show that cos𝜽 + 1 -sin𝜽 =2 sec𝜽
1-sin𝜽 cos𝜽
sol) cross multiply we get,
cos^2 𝜽 + (1-sin)^2 𝜽
(1-sin𝜽)(cos𝜽)
***********************************
(1-sin)^2 is in (a-b)^2 form expand we get
************************************
cos^2 𝜽 + (1 + sin^2𝜽 -2*1*sin𝜽)
(1- sin𝜽) cos𝜽
**********************
WE KNOW,
sin^2 𝜽 + cos^2 𝜽 = 1
***********************
(Sin^2 𝜽 + cos^2 𝜽) + 1 - 2sin𝜽
(1-sin𝜽) cos𝜽
1+ 1- 2sin𝜽)
(1-sin𝜽) cos𝜽
2-2 sin𝜽
(1-sin𝜽) cos𝜽
taking"2" common from numerator
2(1-sin𝜽)
2
cos𝜽
2 * 1
cos𝜽
From Trigonometry identities we know ( 1 /cos𝜽 = sec𝜽)
.^. 2 sec𝜽
**************************************************
15(b) In a right angled triangle the hypotenuse is 10cm more than the shorter side. If third side is 6cm less than hypotenuse, find the sides of right angled triangle
sol)
condition 1) :- Hyp is 10cm more than shorter side
Let "x" be the shorter side
.^. Hyp = 10+x
cond 2) :- 3rd side 6cm less than Hyp
Hyp - 6cm
10+x-6 = x + 4 cm
By using Pythagoras theorem
AC^2 = AB^2 + BC^2
( x+10)^2 = ( x)^2 + ( x+4)^2
**********************************
(x+10)^2 and (x+4)^2 is in (a+b)^2 form
***************************************************
x^2 + 100 +20x = x^2 + ( x^2 +16+8x)
100 + 20x = x^2 + 16 + 8x
84 + 20x = x^2 + 8x
x^2 - 12x - 84 = 0
we have
x = -b + √(b^2 - 4ac)
2a
x = -(-12) + √(-12)^2 - 4 * 1 * (-84)
2 * 1
x = 12 + √144 + 336
2
x = 12 + √480
2
x = 12 + 21.90
2
x = (12 + 21.90)
2
=> (33.90)/2 = 16.95
or
x = 12 - 21.90
2
=> -4.95
As sides cannot b negative we take x= 19.95
now
Shorter side (x) = 16.95 cm
Third side (x + 4) = 20.95 cm
Hyp (x + 10) = 26.95 cm
***********************************************
16) Find the mean age of 100 residents of a colony from the following data
sol) Assumed Mean(A) = 35
𝝨 fi = 10 + 15 + 25 + 25+10+10+5 = 100
𝝨 fi*Ui = -30-30-25+0+10+20+15 = -40
Mean:-
Assumed meand(A) = 35
Class interval(h) = (20-10) = 10
𝚺fi = 100
𝚺fi * Ui = -40
X = a + h [ 𝞢 fi ui]
𝞢 fi
=> 35 + 10 [ -40 ]
100
=> 35 +(-4)
=35-4
=31 is the mean age of 100 residents
*********************************************
16(b) A toy is made with seven equal cubes of side √7 cm. six cubes are joined to six faces of a seventh cube find the total surface of the toy.
sol) Given :- No.of cube = 7
Side of the cube = s = √7 cm
Six cubes are joined to 6 faces of a 7th cube
if one cube attached with other cube one of its face get hidden and only five faces of cube are exposed and only surface area of six cube is effective
we have
Surface area of one side cube = S^2 = (√7)^2 = 7 cm^2
.^. , surface area of five side of a cube = 5 * 7 = 35 cm^2
.^. surface area of six cube = 6 * 35 = 210 cm^2
Total surface area of the toy = 210 cm^2
***************************************************
17(a) If two dice are thrown at the same time, find the probability of getting sum of the dots on top is prime.
sol) Prime Number:- which is divisible by 1 and itself
2,3,5,7,11,13, 15........
(1,1) ,( 1,2),(1,4), (1,6) = 4
(2,1),(2,3),(2,5) =3
(3,2), (3,4) =2
(4,1), (4,3) =2
(5,2), (5,6) =2
(6,1),(6,5) =2
----------------------
Total 15 outcomes
.^. probability for getting sum as prime number
=> No.of favourable event
Total No.of outcomes
=> 15
36
=> 5 is the required probability
12
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17 (b) The angle of elevation of the top of a hill at the foot of a tower is 60 degree and the angle of elevation of the top of the tower from the foot of the hill is 30 deg. If the tower is 50m high, what is the height of the hill.
sol)
(!) In ⧍ ABC
AB = Tan 60
BC
50 = 1
BC √3
BC = 50 * √3 ---------------------(1)
(!!) In ⧍ DBC
DC = Tan 30 deg
BC
DC = √3
50√3 -------------->(From (1))
DC = 50√3 * √3
DC = 50 * 3
DC = 150 m is the height of the hill
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