CBSE Maths Test Paper 2019-Class X
General Instructions:
(i) This question paper comprises four sections – A, B, C and D. This question paper carries 40 questions. All questions are compulsory.
(ii) Section A: Q. No.1 to 17 comprises of 17 questions of one mark each.
(iii) Section B: Q. No. 18 to 26 comprises of 9 questions of two mark each.
(iv) Section C: Q. No. 27 to 34 comprises of 8 questions of three mark each.
(v) Section D: Q. No. 35 to 40 comprises of 6 questions of four mark each.
(vi) Use of Calculators are not permitted.
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Question 1 of 40
1. Question
1 point(s)Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2, -3) and B is the point (1, 4).
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Let the co-ordinates of point A be (x, y) and point O (2, -3) be point the centre, then
By mid-point formula,
and
or and
The co-ordinates of point A are (3, -10) -
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Question 2 of 40
2. Question
1 point(s)For what values of \(k \), the roots of the equation \(x^2+4x+k=0 \) are real?
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The given equation is
On comparing the given equation with, we get and
For real roots,
or
or
or
For, equation will have real roots. -
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Question 3 of 40
3. Question
1 point(s)Find the value of k for which the roots of the equation \(3x^2-10x+k=0 \) are reciprocal of each other.
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The given equation is
On comparing it with, we get
Let the roots of the equation are and
Product of the roots
or -
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Question 4 of 40
4. Question
1 point(s)Find \(A \) if \(\tan2A=\cot(A-24^o) \)
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Given,
or
or
or
or -
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Question 5 of 40
5. Question
1 point(s)Find the value of \((\sin^2\ 33^o+\sin^2\ 57^o) \)
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-
-
Question 6 of 40
6. Question
1 point(s)How many two digits numbers are divisible by 3?
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The two-digit numbers divisible by
are
This is an A.P. in which
or
or
So, there are two-digit numbers divisible by . -
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Question 7 of 40
7. Question
1 point(s)In Figure below, \(DE || BC, AD = 1\ cm \) and \(BD = 2\ cm \). What is the ratio of the ar \((\triangle ABC) \) to the ar \((\triangle ADE) \)?s
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Given,
Also,
(Given)Â (corresponding angles)In
and (common) [by equation (i)] (by AA rule)Now,
-
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Question 8 of 40
8. Question
1 point(s)Find a rational number between \(\sqrt2 \) and \(\sqrt3 \).
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As
So, a rational number between and is -
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Question 9 of 40
9. Question
1 point(s)Find the coordinates of a point A, where AB is a diameter of the circle with centre (-2, 2) and B is the point with coordinates (3, 4).
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By mid-point formula
and
Co-ordinates of point A are. -
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Question 10 of 40
10. Question
1 point(s)Find the value of \(k \) for which the following pair of linear equations have infinitely many solutions.
\(2x + 3y = 7, (k + 1)x + (2k – 1)y = 4k + 1 \)-
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Given,
and
On comparing above equations with and , we get
For infinitely many solutions
or
Hence,. -
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Question 11 of 40
11. Question
1 point(s)Two positive integers a and b can be written as \(a=x^3y^2 \) and \(b=xy^3. \ x,y \) are prime numbers. Find LCM \((a,b) \).
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Given,
and
L.C.M = Product of the greatest power of each prime factors = -
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Question 12 of 40
12. Question
1 point(s)Find, how many two-digit natural numbers are divisible by 7.
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Two digit numbers which are divisible by 7 are 14, 21, 28,…. 98
Hence, there are 13 two digit numbers, divisible by 7. -
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Question 13 of 40
13. Question
1 point(s)If the sum of first n terms of an A.P. is \(n^2 \), then find its 10th term.
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Given,
Put
Hence 10th term -
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Question 14 of 40
14. Question
1 point(s)Find the HCF of 1260 and 7344 using Euclid’s algorithm.
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Two numbers are
and
Since, we apply the Euclid division lemma to and , we get
H.C.F. of and is . -
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Question 15 of 40
15. Question
1 point(s)Which term of the A.P. 3, 15, 27, 39, …… will be 120 more than its 21st term?
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The given A.P. is
Here
Now,
Hence, the term which is 120 more than its 21st term will be its 31st term. -
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Question 16 of 40
16. Question
1 point(s)A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.
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When a coin is tossed three times, the set of all possible outcomes is given by,
Same result on all tosses -
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Question 17 of 40
17. Question
1 point(s)A die is thrown once. Find the probability of getting a number which is a prime number
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In throwing a die Total possible outcomes
i.e.,
Prime numbers -
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Question 18 of 40
18. Question
2 point(s)If the common difference of an AP is 3, find \(a_{20}-a_{15} \)
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-
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Question 19 of 40
19. Question
2 point(s)Find the value of \(a \), So that the point \((3,a) \) lies on the line represented by \(2x-3y=5 \)
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-
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Question 20 of 40
20. Question
2 point(s)For what value of \(k \) will \(k+9,2k-1 \) and \(2k+7 \) are the consecutive terms of an AP?
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-
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Question 21 of 40
21. Question
2 point(s)Find \( c\) if the system of equations \(cx + 3y + (3 – c) = 0,\ 12x + cy – c = 0 \) has infinitely many solutions?
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The given equations are
and
On comparing with equation and equation , we get
and
For infinitely many solutions
or or
So, from both the above cases -
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Question 22 of 40
22. Question
2 point(s)Prove that \(\sqrt2 \) is an irrational number.
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Let
is a rational number.
So, where and are co-prime integers and
or
Squaring on both sides, we get
Therefore, divides
or divides a (from theorem)
Let, for some integer
From equation (i)
or
or
It means that divides and so divides
Therefore and have at least as a common factor.
But this contradicts the fact that and are co-prime.
This contradiction is due to our wrong assumption that is rational.
So, we conclude that is irrational.
Hence Proved. -
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Question 23 of 40
23. Question
2 point(s)Find the value of \(k \) such that the polynomial \(x^2-(k+6)x+2(2k-1) \) has sum of its zeros equal to half to their product.
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The given quadratic polynomial is
Comparing with, we get and
Let the zeroes of the polynomial be and
we know that
or
Also,
or
According to question
Sum of zeroes of their product
or [using equations (i) & (ii)]
or -
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Question 24 of 40
24. Question
2 point(s)A father’s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.
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Let the present age of father be
years and sum of ages of his two children be years
According to question
After 5 years
Father’s age years
Sum of ages of two children years years
According to question
or
or
or (Using equations (i))
Now from equation (i) (Put )
or
So, Present age of father years. -
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Question 25 of 40
25. Question
2 point(s)Find the point on y-axis which is equidistant from the points (5, -2) and (-3, 2)
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We know that a point on the
-axis is of the form .
So, let the point be equidistant from and
Then
or
or
or
So, the required point is -
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Question 26 of 40
26. Question
2 point(s)The line segment joining the points \(A(2, 1) \) and \(B(5, -8) \) is trisected at the points \(P \) and \(Q \) such that \( P\) is nearer to \(A \). If \(P \) also lies on the line given by \(2x – y + k = 0 \), find the value of \(k \).
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The line segment
is trisected at the points and and is nearest to
So, divides in the ratio
Then co-ordinates of P, by section formula
lies on the line
On putting and in the given equation, we get
Hence, -
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Question 27 of 40
27. Question
3 point(s)Prove that \((\sin\theta+\text{cosec}\ \theta)^2+(\cos\theta+\sec\theta)^2=7+\tan^2\theta+\cot^2\theta \).
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L.H.S.
. -
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Question 28 of 40
28. Question
3 point(s)Prove that \((1+\cot A-\text{cosec}\ A)\ (1+\tan A+\sec A)=2 \).
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L.H.S.
-
-
Question 29 of 40
29. Question
3 point(s)In Fig. \(PQ \) is a chord of length 8 cm of a circle of radius 5 cm and centre O. The tangents at \(P \) and \(Q \) intersect at point \(T \). Find the length of \(TP \).
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Join
, let it intersect at the point
Now, is an isosceles triangle and is the angle bisector of .
So, and therefore, bisects
Also,
Now,
Hence, the length of -
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Question 30 of 40
30. Question
3 point(s)In Fig. \(\angle ACB=90^o \) and \(CD\bot AB \), prove that \(CD^2=BD\times AD \).
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Given,
in which and
To prove:
Proof: In and
(common)
( each)
(By AA rule)…(i)
Similarly,
(By AA rule)…(ii)
From equation (i) and (ii)
(by the definition of similarity of triangles)
or
or
Hence Proved. -
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Question 31 of 40
31. Question
3 point(s)If \(P \) and \(Q \) are the points on side \(CA \) and \(CB \) respectively of \(\triangle ABC \), right-angled at \( C\), prove that \((AQ^2+BP^2)=(AB^2+PQ^2) \).
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Given,
is a right-angled triangle in which
To prove:
Construction: Join and
Proof: In
(Using Pythagoras theorem)
In
(Using Pythagoras theorem)
Adding equation (i) and (ii)
or
Hence Proved. -
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Question 32 of 40
32. Question
3 point(s)Find the area of the shaded region in Fig. if ABCD is a rectangle with sides 8 cm and 6 cm and D is the centre of the circle.
[Take \( \pi=3.14\)]-
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Given,
is a rectangle with sides and
In
(By Pythagoras Theorem)
The diagonal of the rectangle will be the diameter of the circle radius of the circle
Area of shaded portion = Area of circle – Area of Rectangle
Hence, Area of shaded portion -
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Question 33 of 40
33. Question
3 point(s)Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/hour. How much area will it irrigate in 30 minutes, if 8 cm standing water is needed?
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Let
be the width and be the depth of the canal and
Water is flowing with a speed
Length of water flowing in
Length (l) of water flowing in
Volume of water flowing in 30 min.
Let the area irrigated in 30 min be
Volume of water required for irrigation = Volume of water flowing in 30 min. hectares ( hectares )
Hence, the canal will irrigate hectares in min. -
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Question 34 of 40
34. Question
3 point(s)Find the mode of the following frequency distribution.
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The given frequency distribution table is
Here, the maximum class frequency is
Modal class
lower limit (l) of modal class
Class size (h)
Frequency of the modal class
Frequency of preceding class
Frequency of succeeding class
Hence, Mode. -
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Question 35 of 40
35. Question
4 point(s)Two water taps together can fill a tank in \(1\dfrac{7}{8} \) hours. The tap with longer diameter takes 2 hours less than the tap with a smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.
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Portion of tank filled by A in 1 hour
Portion of tank filled by B in 1 hour
Portion of tank filled by both taps in 1 hour
Time taken with long diameter tap hours
Time taken with smaller diameter tap hours -
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Question 36 of 40
36. Question
4 point(s)If the sum of first four terms of an A.P. is \(40 \) and that of first \(14 \) terms is \(280 \). Find the sum of its first \(n \) terms.
-
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Given,
and
If a be the first term and d be the common difference of an A.P.
Then, Sum of term
or
or
or
or ….(i)
Also, Sum of first terms
or ….(ii)
On solving equation (i) and (ii), we get
Now, sum of terms
On putting
Hence, the sum of first terms is -
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Question 37 of 40
37. Question
4 point(s)A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from \(60^o \) to \(30^o \). Find the speed of the boat in metres per minute. [Use latex]\sqrt3=1.732 [/latex]]
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Let
be the lighthouse and be the two positions of the boat, such that,
and
Now, In
….(i)
In
or ….(ii)
From equation (i) and (ii)
metres
Time taken to cover
Speed of boat
Hence, speed of boat -
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Question 38 of 40
38. Question
4 point(s)Construct a \(\triangle ABC \) in which \(CA=6\ cm,\ AB=5\ cm \) and \(\angle BAC=45^o \). Then construct a triangle whose sides are \(35 \) of the corresponding sides of \(\triangle ABC \).
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Steps of Construction are as follows:
1. Draw
2. At the point, draw .
3. From cut off
4. Join is formed with given data.
5. Draw making an acute angle with as shown in the figure.
6. Draw arcs and with equal intervals.
7. Join.
8. Draw meeting at .
9. From, draw meeting at
Hence is the required triangle. -
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Question 39 of 40
39. Question
4 point(s)A bucket open at the top is in the form of a frustum of a cone with a capacity of \(12308.8\ cm^3 \). The radii of the top and bottom of circular ends of the bucket are \( 20\ cm\) and \(12\ cm \) respectively. Find the height of the bucket and also the area of the metal sheet used in making it. (Use \(\pi=3.14 \))
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Let
and be the radii of the top and the bottom circular ends of the bucket respectively.
Let be the height of the bucket. and
Capacity of the bucket
Volume of bucket (frustum)
or
or
or
Thus, the height of the bucket is.
The area of the metal sheet used in making the bucket = CSA of bucket + area of the circular bottom
Area of metal sheet used -
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Question 40 of 40
40. Question
4 point(s)Prove that in a right-angle triangle, the square of the hypotenuse is equal the sum of squares of the other two sides.
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Given,
right angled at .
To prove:
Construction :
Proof: In and
(common)
( each)
(By AA rule)
So, (sides are proportional)
or
Also, In and
(common)
( each)
So,
or
Adding equation (i) and (ii), we get
or
Hence Proved. -
and 
and 
, we get
and 
will have real roots.
and 
![\cot(90^o-2A)=\cot(A-24^o)\ [\because\tan\theta=\cot(90^o-\theta)]](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-04637e40b3fd39db8212c89641a85f60_l3.png "Rendered by QuickLaTeX.com")
![\sin^2\ 33^o+\sin^2\ 57^o\\ \\ =\sin^2\ 33^o+\cos^2\ (90^o-57^o)\\ \\ =\sin^2\ 33^o+\cos^2\ 33^o\\ \\ =1[\therefore\sin^2\theta+\cos^2\theta=1]](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-1d4dbac66cf8fcaa64f0643108324ad8_l3.png "Rendered by QuickLaTeX.com")
are 
two-digit numbers divisible by 
(Given)Â 
(corresponding angles)
and 
(common)
[by equation (i)]
(by AA rule)![\dfrac{\text{ar}\left(\triangle ABC\right)}{\text{ar}\left(\triangle ADE\right)}=\left(\dfrac{AB}{AD}\right)^2\\ \\ \left[\because\triangle ABC\sim\triangle ADE\right]\\ \\ \text{or} \ \dfrac{\text{ar}\left(\triangle ABC\right)}{\text{ar}\left(\triangle ADE\right)}=\left(\dfrac{3}{1}\right)^2=\dfrac{9}{1}\\ \\ \therefore\text{ar}\ \left(\triangle ABC\right):\text{ar}\ \left(\triangle ADE\right)=9:1](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-4b8ec824ac8131a81f0f806a1eccbaf3_l3.png "Rendered by QuickLaTeX.com")
and 
is 
.
and 
and 
, we get
.
and 
= Product of the greatest power of each prime factors = 
![a_n=S_n-S_{n-1}\\ \\ =n^2-(n-1)^2\\ \\ =n^2-[n^2-2n+1]\\ \\ =n^2-n^2+2n-1\\ \\ a_n=2n-1](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-2ac6b37c70cf718da52bdffcf140d374_l3.png "Rendered by QuickLaTeX.com")
and 
, we apply the Euclid division lemma to 
.
![a_{20}-a_{15}=[a+19d]-[a+14d]\\ \\ =19d-14d\\ \\ =5d=15](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-28eb3a19005e65cf8d38ef40244bfd40_l3.png "Rendered by QuickLaTeX.com")
and equation 
, we get
or 
where 
and 
are co-prime integers and 
divides 
, for some integer 
and so 
and 
, we get 
and 
and 
of their product
[using equations (i) & (ii)]
years and sum of ages of his two children be 
years
years
years 
years
(Using equations (i))
(Put 
years.
.
be equidistant from 
and 
is trisected at the points 
and 
and 
![=\mathrm{P}\left[\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\right]\\ \\ =\mathrm{P}\left[\frac{1(5)+2(2)}{1+2}, \frac{1(-8)+2(1)}{1+2}\right]\\ \\ =\mathrm{P}\left[\frac{5+4}{2}, \frac{-8+2}{3}\right]=\mathrm{P}(3,-2)](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-be1208b969f091a71956959a2c4b68eb_l3.png "Rendered by QuickLaTeX.com")
and 
in the given equation, we get
.![=\sin ^{2} \theta+\cos ^{2} \theta+\operatorname{cosec}^{2} \theta+\sec ^{2} \theta+2 \sin \theta \cdot \frac{1}{\sin \theta}+2 \cdot \cos \theta \cdot \frac{1}{\cos \theta}\\ \\ \left[\because \operatorname{cosec} \theta=\frac{1}{\sin \theta}\ \sec \theta=\frac{1}{\cos \theta}\right]\\ \\ =1+1+\cot ^{2} \theta+1+\tan ^{2} \theta+4\\ \\ \left[\because \operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta \sec ^{2} \theta=1+\tan ^{2} \theta\right]\\ \\ =7+\tan^2\theta+\cot^2\theta\ \text{(R.H.S) Hence Proved}.](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-973849bff372ca902982556d4fa4711d_l3.png "Rendered by QuickLaTeX.com")
![=\left(1+\dfrac{\cos A}{\sin A}-\dfrac{1}{\sin A}\right)\left(1+\dfrac{\sin A}{\cos A}+\dfrac{1}{\cos A}\right)\\ \\ =\left(\dfrac{\sin A+\cos A-1}{\sin A}\right)\left(\dfrac{\cos A+\sin A+1}{\cos A}\right)\\ \\ \left[\because(a+b)(a-b)=a^{2}-b^{2}\right]\\ \\ =\dfrac{(\sin A+\cos A)^{2}-1}{\sin A \cdot \cos A}\\ \\ =\dfrac{\sin ^{2} A+\cos ^{2} A+2 \cdot \sin A \cdot \cos A-1}{\sin A \cdot \cos A}\\ \\ =\dfrac{1+2 \sin A \cdot \cos A-1}{\sin A \cdot \cos A}=\dfrac{2 \sin A \cdot \cos A}{\sin A \cdot \cos A} \\ \\ =2\ \text{(R.H.S.)}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-9d69f9c997c87a2740703e3dba8d5449_l3.png "Rendered by QuickLaTeX.com")
, let it intersect 
at the point 
is an isosceles triangle and 
is the angle bisector of 
.
and therefore, 
![[\because\ \text{In}\ \triangle TRP,\angle TRP=90^o]\\ \\ \Rightarrow\ \angle RPO=\angle PTR\\ \\ \text{So,}\ \triangle TRP\sim\triangle PRO\ \text{(By AA rule)}\\ \\ \therefore\ \dfrac{TP}{PO}=\dfrac{RP}{RO}\\ \\ \text{or}\ \dfrac{TP}{5}=\dfrac{4}{3},\ \text{or}\ TP=\dfrac{20}{3}\ cm](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-4b6750ce6be1ef39e7a55cda0d328070_l3.png "Rendered by QuickLaTeX.com")
in which 
and 
and 
(
each)
(By AA rule)…(i)
(By AA rule)…(ii)
is a right-angled triangle in which 
and 
(Using Pythagoras theorem)
(Using Pythagoras theorem)
is a rectangle with sides 
and 
(By Pythagoras Theorem)
be the depth of the canal
and 
be 
hectares (
hectares 
)
hectares in 
of the modal class 
of preceding class 
of succeeding class 
.
hours
hours
and 
term ![(S_n)=\dfrac{n}{2}[2a+(n-1)d]](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-722cff3e87d3284f32fc07c2424e316e_l3.png "Rendered by QuickLaTeX.com")
![S_n=\dfrac{n}{2}[2a+(n-1)d]](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-d4cb5bf3e2d7bc08f09c09bb7f552a3f_l3.png "Rendered by QuickLaTeX.com")
![S_4=\dfrac{4}{2}[2a+3d]](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-2054c8882d948c3663c0504e58db8204_l3.png "Rendered by QuickLaTeX.com")
….(i)
terms 
![S_{14}=\dfrac{14}{2}[2a+(13)d]\\ \\ 280=7(2a+13d)](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-428d1e8a306441d77f691d4242ec4aab_l3.png "Rendered by QuickLaTeX.com")
….(ii)
terms![S_n=\dfrac{n}{2}[14+(n-1)2]\\ \\ =n[7+n-1]\\ \\ =n(n+6)\\ \\ =n^2+6n](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-0a87e49ff4d9ce7fb9fda911b0f0f621_l3.png "Rendered by QuickLaTeX.com")
and 
be the two positions of the boat, such that,
and 
….(i)
….(ii)
metres
Time taken to cover 
Speed of boat 
.
cut off 
is formed with given data.
making an acute angle with 
arcs 
and 
with equal intervals.
.
meeting 
.
, draw 
meeting 
at 
is the required triangle.
and 
be the radii of the top and the bottom circular ends of the bucket respectively.
and 
.
![=\pi[(R+r)l + r^2]](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-3f86cb54ba9d2fbadf52daa677437729_l3.png "Rendered by QuickLaTeX.com")
![= 3.14 [(20 + 12) \times 17 + 12^2]\\ \\ = 3.14 [32 \times 17 + 144]\\ \\ = 3.14 [544 + 144]\\ \\ = 3.14 \times 688\\ \\ = 2160.32\ cm^2](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-905e12f08096b0a380ac7c240779fdd0_l3.png "Rendered by QuickLaTeX.com")
right angled at 
.
and 
(
(By AA rule)
(sides are proportional)
and 
(common)
(
