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You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Additional Questions

EXERCISE 9.1

Question 1.
An unbiased die is thrown. What is the probability of getting
(i) an even number or a multiple of 3.
(ii) a number between 3 and 6.
Solution:
(i) Probability of getting an even number \(\frac{3}{6}=\frac{1}{2}\)
Probability of getting a multiple of 3 = \(\frac{2}{6}\)
Probability of getting an even multiple of 3 = \(\frac{1}{6}\)
Probability of getting an even number or


Question 2.
Two unbiased coins are tossed simultaneously find the probability of getting
(i) two heads
(ii) one head
(iii) at least one head
(iv) at most one head.
Solution:
S = {HH, HT, TH, TT}
(i) probability of two heads = \(\frac{1}{4}\)
(ii) probability of one head = \(\frac{1}{2}\)
(iii) probability of at least one head = \(\frac{3}{4}\)
(iv) probability of at most one head = \(\frac{3}{4}\)

Question 3.
Find the probability that a leap year selected at random will contain 53 Sundays.
Solution:
S = {Sunday Monday, Monday Tuesday, Tuesday Wednesday, Wednesday Thursday, Thursday Friday, Friday Saturday, Saturday Sunday}
n (S) = 7; n (A) = 2; P(A) = \(\frac{2}{7}\)

Question 4.
What is the probability that a number selected from the numbers 1, 2, 3, 25 is a prime number when each of the given numbers is equally likely to be selected?
Solution:
A = {2, 3, 5, 7, 11, 13, 17, 19,23}
P(A) = \(\frac{9}{25}\)

EXERCISE 9.2

Question 1.
Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?
Solution:
A = {3, 6, 9, 12, 15, 18, 7, 14}

Question 2.
One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is
(i) an ace,
(ii) either red card or king.
Solution:

Question 3.
A bag contains 3 red and 2 blue marbles. A marble is drawn at random. What is the probability of drawing a blue marble?
Solution:

Question 4.
Two dice are thrown simultaneously. Find the probability of getting
(i) an even number as the sum.
(ii) a total of at least 10
(iii) a doublet of even number.
Solution:

n(S) = 36


Question 5.
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Solution:

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Ex 9.3

Question 1.
A number between 0 and 1 that is used to measure uncertainty is called
(1) Random variable
(2) Trial
(3) Simple event
(4) Probability
Solution:
(4) Probability

Question 2.
Probability lies between
(1) -1 and +1
(2) 0 and 1
(3) 0 and n
(4) 0 and ∞
Solution:
(2) 0 and 1

Question 3.
The probability based on the concept of relative frequency theory is called
(1) Empirical Probability
(2) Classical Probability
(3) Both (1) and (2)
(4) Neither (1) or (2)
Solution:
(1) Empirical Probability

Question 4.
The probability of an event cannot be
(1) Equal to zero
(2) Greater than zero
(3) Equal to one
(4) Less than zero
Solution:
(4) Less than zero

Question 5.
The probability of all possible outcomes of a random experiment is always equal to
(1) one
(2) Zero
(3) Infinity
(4) Less than one
Solution:
(1) one

Question 6.
If A is any event in S then its complement is A’ then, P(A’) is equal to
(1) 1
(2) 0
(3) 1 – A
(4) 1 – P(A)
Solution:
(4) 1 – P(A)

Question 7.
Which of the following cannot be taken as probability of an event?
(1) 0
(2) 0.5
(3) 1
(4) -1
Solution:
(4) -1

Question 8.
A particular result of an experiment is called
(1) Trial
(2) Simple event
(3) Compound event
(4) Outcome
Solution:
(4) Outcome

Question 9.
A collection of one or more outcomes of an experiment is called
(1) Event
(2) Outcome
(3) Sample point
(4) None of above
Solution:
(1) Event

Question 10.
The six faces of the dice are called equally likely if the dice is
(1) Small
(2) Fair
(3) Six-faced
(4) Round
Hint: Fair means all outcomes are equally likely.
Solution:
(2) Fair

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Ex 9.2

Question 1.
A company manufactures 10000 Laptops in 6 months. Out of which 25 of them are found to be defective. When you choose one Laptop from the manufactured, what is the probability that selected Laptop is a good one.
Solution:
Total n(S) = 10,000
Defective n( A) = 25

No. of good laptops = 1000 – 25
n(B) = 9975


Question 2.
In a survey of 400 youngsters aged 16-20 years, it was found that 191 have their voter ID card. If a youngster is selected at random, find the probability that the youngster does not have their voter ID card.
Solution:
No. of youngsters n(S) = 400
No. of youngsters having voter id n(A) = 191
No. of youngsters do not have their voter id n(B) = 400 – 191 = 209

Question 3.
The probability of guessing the correct answer to a certain question is \(\frac{x}{3}\). If the probability of not guessing the correct answer is \(\frac{x}{5}\), then find the value of x.
Solution:

Question 4.
If a probability of a player winning a particular tennis match is 0.72. What is the probability of the player loosing the match?
Solution:
P(A) = 0.72
P(A’) = 1 – 0.72 = 0.28

Question 5.
1500 families were surveyed and following data was recorded about their maids at homes

A family is selected at random. Find the probability that the family selected has
(i) Both types of maids
(ii) Part time maids
(iii) No maids
Solution:
n(S) = 1500 (Total families)
n(A) = 860 (Part time maids)
n(B) = 370 (Only full time)
n(A∩B) = 250 (Both)


Total families n(S) = 1500
No. of families have maids = 860 + 250 + 370 = 1480
No. of families do not have maids = 1500 – 1480
n( A) = 20

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions

Question 1.
Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
Solution:
Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).
We are given that AP = BP. So, AP² = BP²
(x – 7)² + (y – 1)² =(x – 3)² + (y – 5)²
x² – 14x + 49 + y² – 2y + 1 = x² – 6x + 9 + y² – 10y + 25
x – y = 2
Which is the required relation

Question 2.
Show that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a square.
Solution:
Let A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) be the given points. To prove that ABCD is a square, we have to prove that all its sides are equal and both its diagonals are equal.

Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Therefore, ABCD is a square.

Question 3.
If A (-5, 7), B (-4, -5), C (-1, -6) and D (4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.
Solution:
By joining B to D, you will get two triangles ABD and BCD.
Now, the area of ∆ABD

So, the area of quadrilateral ABCD = 53 + 19 = 72 square units.

Question 4.
Find the coordinates of the points of trisection (i.e. points dividing in three equal parts) of the line segment joining the points A(2, -2) and B(-7, 4).
Solution:
Let P and Q be the points of trisection at AB. i.e., AP = PQ = QB

Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates at P, by applying the section formula, are

Now, Q also divides AB internally in the ratio 2 : 1, so, the coordinates at Q are

Therefore, the coordinates of the points at trisection of the line segment joining A and B are (-1, 0) and (-4, 2).

Question 5.
If the points A(6, 1), B(8, 2), C(9, 4) and D(P, 3) are the vertices of a parallelogram, taken in order. Find the value of P.
Solution:
We know that diagonals of a parallelogram bisect each other.
So, the coordinates at the mid-point of AC = coordinates of the mid-point of BD.

Question 6.
Find the area of a triangle whose vertices are (1,-1), (-4, 6) and (-3, -5).
Solution:
The area of the triangle formed by the vertices A(1, -1), B(-4, 6) and C(-3, -5), by using the formula

So, the area of the triangle is 24 square units.

Question 7.
If A(-2, -1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b.
Solution:
We know that the diagonals of a parallelogram bisect each other. Therefore the co-ordinates of the midpoint of AC are same as the co-ordinates of the mid-point of BD. i.e.

Question 8.
Find the area of the quadrilateral whose vertices, taken in order, are (-3, 2), (5, 4), (7, -6) and (-5, -4).
Solution:
We have Area of the quadrilateral

Question 9.
Find the area of the triangle formed by the points P(-1.5, 3), Q(6, -2) and R(-3, 4).
Solution:
The area of the triangle formed by the given points is equal to

Can we have a triangle of area 0 square units? What does this mean?
If the area of a triangle is 0 square units, then its vertices will be collinear.

Question 10.
Find the value of k if the pointsA(2, 3), B(4, k) and (6, -3) are collinear.
Since the given points are collinear, the area a the triangle formed by them must be 0, i.e.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Unit Exercise 6

Question 1.
Prove that

Solution:

Question 2.

Solution:

Hence proved

Question 3.
If x sin³θ + y cos³θ = sin θ cos θ and x sin θ =
y cos θ , then prove that x² + y² = 1.
Solution:
x sin³θ +y cos³θ= sinθ cosθ ; x sinθ y cosθ.
x (sinθ) [sin²θ + cos²θ] = sinθ cosθ

Question 4.
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\)
Solution:

Hence Proved.

Question 5.
A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such away that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Determine the speed at which the bird flies. ( \(\sqrt{3}\) = 1.732).
Solution:
Let s be the speed of the bird. In 2 seconds, the bird goes from C to D, it covers a distance ‘d’

Question 6.
An aeroplane is flying parallel to the Earth’s surface at a speed of 175 m/sec and at a height of 600 m. The angle of elevation of the aeroplane from a point on the Earth’s surface is 37° at a given point. After what period of time does the angle of elevation increase to 53°? (tan 53° = 1.3270, tan 37° = 0.7536)
Solution:
Let Plane’s initial position be A. Plane’s final position = D Plane travels from A ➝ D.

Question 7.
A bird is flying from A towards B at an angle of 35°, a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48° and distance 32 km away.
(i) How far is B to the North of A?
(ii) How far is B to the West of A?
(iii) How far is C to the North of B?
(iv) How far is C to the East of B?
(sin 55° = 0.8192, cos 55° = 0.5736,
sin 42° = 0.6691, cos 42° = 0.7431)
Solution:

Question 8.
Two ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45° respectively. If the
distance between the ships is \(200\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)\) metres, find the height of the lighthouse.
Solution:
From the figure AB – height of the light house = h CD – Distance between the ships


∴ The height of the light house is 200 metres.

Question 9.
A building and a statue are in opposite side of a street from each other 35 m apart. From a point on the roof of building the angle of elevation of the top of statue is 24° and the angle of depression of base of the statue is 34°. Find the height of the statue.
(tan 24° = 0.4452, tan 34° = 0.6745)
Solution:

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Unit Exercise 5

Question 1.
PQRS is a rectangle formed by joining the points P(-1, -1), Q(-1, 4) ,R(5, 4) and S(5, -1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.
Solution:
A, B, C and D are mid points of PQ, QR, RS & SP respectively.


∴ AB and BC are not perpendicular
⇒ ABCD is rhombus as diagonals are perpendicular and sides are not perpendicular.

Question 2.
The area of a triangle is 5 sq.units. Two of its vertices are (2, 1) and (3, -2). The third Vertex is (x, y) where y = x + 3 . Find the coordinates of the third vertex.
Solution:
Area of triangle formed by points (x₁, y₁),

Question 3.
Find the area of a triangle formed by the lines 3x + y – 2 = 0, 5x + 2y – 3 = 0 and 2x – y – 3 = 0
Solution:

Question 4.
If vertices of a quadrilateral are at A(-5, 7), B(-4, k) , C(-1, -6) and D(4, 5) and its area is
72 sq.units. Find the value of k.
Area (quadrilateral ABCD)

Question 5.
Without using distance formula, show that the points (-2, -1) , (4, 0) , (3, 3) and (-3, 2) are vertices of a parallelogram.
Solution:

Slope of AB = Slope of CD
Slope of BC = Slope of DA
Hence ABCD forms a parallelogram.

Question 6.
Find the equations of the lines, whose sum and product of intercepts are 1 and -6 respectively.
Let the intercepts be x₁, y₁ respectively

Question 7.
The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹ 14/litre and 1220 litres of milk each week at ₹ 16 litre. Assuming a linear relationship
between selling price and demand, how many litres could he sell weekly at ₹ 17/litre?
Solution:

Question 8.
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Solution:

Question 9.
Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Solution:
4x + 7y – 3 = 0
2x – 3y + 1 = 0
4x + 7y – 3 – 2(2x – 3y + 1) = 0

Question 10.
A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.
Solution:

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Additional Questions

Question 1.
In figure if PQ || RS, Prove that ∆POQ ~ ∆SOR
Solution:
PQ || RS

So, ∠P = ∠S (A Hernate angles)
and ∠Q = ∠R
Also, ∠POQ = ∠SOR (vertically opposite angle)
∴ ∆POQ ~ ∆SOR (AAA similarity criterion)

Question 2.
In figure OA . OB = OC . OD Show that ∠A = ∠C and ∠B = ∠D
Solution:
OA . OB = OC . OD (Given)

Also we have ∠AOD = ∠COB
(vertically opposite angles) …………. (2)
From (1) and (2)
∴ ∆AOD ~ ∆COB (SAS similarity criterion)
So, ∠A = ∠C and ∠B = ∠D
(corresponding angles of similar triangles)

Question 3.
In figure the line segment XY is parallel to side AC of ∆ABC and it divides the triangle into two parts of equal areas. Find the ratio \(\frac{\mathbf{A X}}{\mathbf{A B}}\)
Solution:
Given XY||AC

Question 4.
In AD⊥BC, prove that AB² + CD² = BD² + AC².
Solution:
From ∆ADC, we have
AC² = AD² + CD² …………… (1)
(Pythagoras theorem)
From ∆ADB, we have
AB² = AD² + BD² …………. (2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB² – AC² = BD² – CD²
AB² + CD² = BD² + AC²

Question 5.
BL and CM are medians of a triangle ABC right angled at A.
Prove that 4(BL² + CM²) = 5BC².
Solution:
BL and CM are medians at the ∆ABC in which
A = ∠90°.
From ∆ABC
BC² = AB² + AC² …………… (1)
(Pythagoras theorem)

4CM² = 4AC² + AB² …………… (3)
Adding (2) and (3), we have
4(BL² + CM²) = 5(AC² + AB²)
4(BL² + CM²) = 5BC² [From (1)]

Question 6.
Prove that in a right triangle, the square of the hypotenure is equal to the sum of the squares of the others two sides.
Solution:
Proof:

We are given a right triangle ABC right angled at B.
We need to prove that AC² = AB² + BC²
Let us draw BD ⊥ AC
Now, ∆ADB ~ ∆ABC
\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AB}}{\mathrm{AC}}\) (sides are proportional)
AD . AC = AB² …………. (1)
Also, ∆BDC ~ ∆ABC
\(\frac{C D}{B C}=\frac{B C}{A C}\)
CD . AC = BC² …………. (2)
Adding (1) and (2)
AD . AC + CD . AC = AB² + BC²
AC(AD + CD) = AB² + BC²
AC . AC = AB² + BC²
AC² = AB² + BC²

Question 7.
A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.
Solution:
Let AB be the ladder and CA be the wall with the window at A.

Also, BC = 2.5 m and CA = 6 m
From Pythagoras theorem,
AB² = BC² + CA²
= (2.5)² + (6)²
= 42.25
AB = 6.5
Thus, length at the ladder is 6.5 m.

Question 8.
In figure O is any point inside a rectangle ABCD. Prove that OB² + OD² = OA² + OC².
Solution:
Through O, draw PQ||BC so that P lies on AB and Q lies on DC.
Now, PQ||BC
PQ⊥AB and PQ⊥DC (∵ ∠B = 90° and ∠C = 90°)
So, ∠BPQ = 90° and ∠CQP = 90°
Therefore BPQC and APQD are both rectangles.
Now from ∆OPB,
OB² = BP² + OP² ……………. (1)
Similarly from ∆OQD,
OD² = OQ² + DQ² …………. (2)
From ∆OQC, we have
OC² = OQ² + CQ² ……….. (3)
∆OAP, we have
OA² = AP² + OP² …………. (4)
Adding (1) and (2)
OB² + OD² = BP² + OP² + OQ² + DQ² (As BP = CQ and DQ = AP)
= CQ² + OP² + OQ² + AP²
= CQ² + OQ² + OP² + AP²
= OC² + OA² [From (3) and (4)]

Question 9.
In ∠ACD = 90° and CD⊥AB. Prove that \(\frac{\mathbf{B C}^{2}}{\mathbf{A C}^{2}}=\frac{\mathbf{B D}}{\mathbf{A D}}\)
Solution:
∆ACD ~ ∆ABC
So,

Question 10.
The perpendicular from A on side BC at a ∆ABC intersects BC at D such that DB = 3 CD. Prove that 2AB² = 2AC² + BC².
Solution:
We have DB = 3 CD

Since ∆ABD is a right triangle (i) right angled at D
AB² – AD² + BD² …………. (ii)
By ∆ACD is a right triangle right angled at D
AC² = AD² + CD² ………… (iii)
Subtracting equation (iii) from equation (ii),
we got

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.5

Question 1.
The area of triangle formed by the points (-5, 0), (0, -5) and (5, 0) is
(1) 0 sq. units
(2) 25 sq.units
(3) 5 sq. units
(4) none of these
Solution:
(2) 25 sq. units
Hint:

Question 2.
A man walks near a wall, such that the distance between him and the wall is 10 units.
Consider the wall to be the Y axis. The path travelled by the man is
(1) x = 10
(2) y = 10
(3) x = 0
(4) y = 0
Solution:
(1) x = 10
Hint:

Question 3.
The straight line given by the equation x = 11 is
(1) parallel to X axis
(2) parallel to Y axis
(3) passing through the origin
(4) passing through the point (0,11)
Solution:
(2) Parallel to y axis
Hint:

Question 4.
If (5, 7), (3, p) and (6, 6) are collinear, then the value of p is
(1) 3
(2) 6
(3) 9
(4) 12
Solution:
(3) 9
If (5, 7), (3, p) and (6, 6) are collinear ∆ = 0

Question 5.
The point of intersection of 3x – y = 4 and x + y = 8 is
(1) (5, 3)
(2) (2, 4)
(3) (3, 5)
(4) (4, 4)
Solution:
(3) (3, 5)]
Hint:

∴ Point of intersection is (3, 5)

Question 6.
The slope of the line joining (12, 3), (4, a) is \(\frac{1}{8}\). The value of ‘a’ is
(1) 1
(2) 4
(3) -5
(4) 2
Solution:
(4) 2
Hint:

Question 7.
The slope of the line which is perpendicular to a line joining the points (0, 0) and (-8, 8) is
(1) -1
(2) 1
(3) \(\frac{1}{3}\)
(4) -8
Solution:
(2) 1
Hint:
Slope of the line joining the points (0, 0) and (-8, 8) is

Question 8.
If slope of the line PQ is \(\frac{1}{\sqrt{3}}\) then slope of the perpendicular bisector of PQ is
(1) \(\sqrt{3}\)
(2) \(-\sqrt{3}(3)\)
(3) \(\frac{1}{\sqrt{3}}\)
(4) 0
Solution:
(2) \(-\sqrt{3}(3)\)

Question 9.
If A is a point on the Y axis whose ordinate is 8 and B is a point on the X axis whose abscissa is 5 then the equation of the line AB is
(1) 8x + 5y = 40
(2) 8x – 5y = 40
(3) x = 8
(4) y = 5
Solution:
(1) 8x + 5y = 40
Hint:

Question 10.
The equation of a line passing through the origin and perpendicular to the line 7x – 3y + 4 = 0 is
(1) 7x – 3y + 4 = 0
(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
Solution:
(3) 3x + 7y = 0
Hint:

Question 11.
Consider four straight lines
(i) l₁ : 3y = 4x + 5
(ii) l₂: 4y = 3x – 1
(iii) l₃: 4y = 3x = 7
(iv) l₄ : 4x + 3y = 2
Which of the following statement is true ?
(1) l₁ and l₂ are perpendicular
(2) l₁ and l₄ are parallel
(3) l₂ and l₄ are perpendicular
(4) l₂ and l₃ are parallel
Solution:
(3) l₂ and l₄ are perpendicular
Hint:

Question 12.
A straight line has equation 8y = 4x + 21. Which of the following is true
(1) The slope is 0.5 and the y intercept is 2.6
(2) The slope is 5 and they intercept is 1.6
(3) The slope is 0.5 and the y intercept is 1.6
(4) The slope is 5 and they intercept is 2.6
Solution:
(1) The slope is 0.5 and the y intercept is 2.6
Hint:

Question 13.
When proving that a quadrilateral is a trapezium, it is necessary to show ………………
(1) Two sides are parallel.
(2) Two parallel and two non-parallel sides.
(3) Opposite sides are parallel.
(4) All sides are of equal length.
Answer:
(2) Two parallel and two non-parallel sides.

Question 14.
When proving that a quadrilateral is a parallelogram by using slopes you must find
(1) The slopes of two sides
(2) The slopes of two pair of opposite sides
(3) The lengths of all sides
(4) Both the lengths and slopes of two sides
Solution:
(2) The slopes of two pair of opposite sides

Question 15.
(2, 1) is the point of intersection of two lines.
(1) x – y – 3 = 0; 3x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
(3) 3x + y = 3; x + y = 7
(4) x + 3y – 3 = 0; x – y – 7 = 0
Answer:
(2) x + y = 3; 3x + y = 7
Hint:
Substitute the value of x = 2 and y = 1 in the given equation.
(1) ⇒ x – y – 3 = 0
2 – 1 – 3 = 0
2 – 4 = 0
– 2 ≠ 0
not true

3x – y – 7 = 0
3(2) – 1 – 7 = 0
6 – 8 = 0
-2 ≠ 0
not true

(2) ⇒ x + y = 3
2 + 1 = 3
3 = 3
True

3x + y = 7
3(2) + 1 = 7
6 + 1 = 7
7 = 7
True
∴ (2, 1) is the point of intersection

(3) ⇒ 3x + y = 3
3(2) + 1 = 3
6 + 1 = 3
7 = 3
not true

x + y = 7
2 + 1 = 7
3 = 7
not true

(4) ⇒ x + 3y – 3 = 0
2 + 3 – 3 = 0
5 – 3 = 0
2 ≠ 0
not true

x – y – 7 = 0
2 – 1 – 7 = 0
2 – 8 = 0
-6 ≠ 0
not true

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.5

Multiple choice questions.
Question 1.
The value of \(\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}\) is equal to
(1) tan²θ
(2) 1
(3) cot²θ
(4) 0
Solution:
(2) 1
Hint:

Question 2.
tan θ cosec²θ – tan θ is equal to
(1) sec θ
(2) cot²θ
(3) sin θ
(4) cot θ
Solution:
(4) cot θ
Hint:

Question 3.
If (sin α + cosec α)² + (cos α + sec α)² = k + tan² α + cot² α, then the value of k is equal to
(1) 9
(2) 7
(3) 5
(4) 3
Solution:
(2) 7
Hint:
(sin α + cosec α)² + (cos α + sec α)² = k + tan² α + cot² α
sin² α + cosec² α + 2 sin α cosec α + cos² α + sec² α + 2 cos α sec α = k + tan² α + cot² α
sin² α + cos² α + cosec² α + sec² α + 2 sin α × \(\frac{1}{\sin \alpha}\) + 2 cos α × \(\frac{1}{\cos \alpha}\) = k + tan² α + cot² α
1 + 1 + cot² α + 1 + tan² α + 2 + 2 = k + tan² α + cot² α
7 + cot² α + tan² α = k + tan² α + cot² α
∴ k = 7

Question 4.
If sin θ + cos θ = a and sec θ + cosec θ = b, then the value of b(a² – 1) is equal to
(1) 2a
(2) 3a
(3) 0
(4) 2ab
Solution:
(1) 2a
a = sin θ + cos θ
b = sec θ + cosec θ

Question 5.
If 3x = sec θ and \(\frac{5}{x}\) = tan θ, then \(x^{2}-\frac{1}{x^{2}}\) is
(1) 25
(2) \(\frac{1}{25}\)
(3) 5
(4) 1
Solution:

Question 6.
If sin θ = cos θ , then 2 tan²θ + sin²θ – 1 is equal to
(1) \(\frac{-3}{2}\)
(2) \(\frac{3}{2}\)
(3) \(\frac{2}{3}\)
(4) \(\frac{-2}{3}\)
Solution:
(2) \(\frac{3}{2}\)
Hint:

Question 7.
If x = a tan θ and y = b sec θ then

Solution:
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
Hint:

Question 8.
(1 + tan θ + sec θ) (1 + cot θ – cosec θ) is equal to
(1) 0
(2) 1
(3) 2
(4) -1
Solution:
(3) 2
Hint:

Question 9.
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p² – q² is equal to ………….
(1) a² – b²
(2) b² – a²
(3) a² + b²
(4) b – a
Answer:
(2) b² – a²
Hint:
p² – q² = (p + q) (p – q)
= (a cot θ + b cosec θ + b cot θ + a cosec θ) (a cot θ + b cosec θ – b cot θ – a cosec θ)
= [cot θ (a + b) + cosec θ (a + b)] [cot θ (a – b) + cosec θ (b – a)]
= (a + b) [cot θ + cosec θ] (a – b) [cosec θ (a – b)]
= (a + b) [cot θ + cosec θ] (a – b) [cot θ – cosec θ]
= (a + b) (a – b) (cot² θ – cosec² θ)
= (a² – b²) (-1) = – (a² – b²)
p² – q² = b² – a^2.

Question 10.
If the ratio of the height of a tower and the length of its shadow is \(\sqrt{3}\) : 1, then the angle of elevation of the sun has measure.
(1) 45°
(2) 30°
(3) 90°
(4) 60°
Solution:
(4) 60°
Hint:

Question 11.
The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘b’ metres above the first, the depression of the foot of the tower is 60°. The height of the tower (in metres) is equal to

Solution:
(2) \(\frac{b}{3}\)
Hint:

Question 12.
A tower is 60 m height. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30°, then x is equal to
(1) 41.92 m
(2) 43.92 m
(3) 43 m
(4) 45.6 m°
Solution:
(2) 43.92 m
Hint:

Question 13.
The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two buildings (in metres) is
(1) 20, 10\(\sqrt{3}\)
(2) 30, 5\(\sqrt{3}\)
(3) 20, 10
(4) 30, 10\(\sqrt{3}\)
Solution:
(4) 30, 10\(\sqrt{3}\)
Hint:


∴ Height of tower = 20 + 10 = 30 m
distance = 17.32 m = 10\(\sqrt{3}\)

Question 14.
Two persons are standing ‘x’ metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is

Solution:
(2) \(\frac{x}{2 \sqrt{2}}\)
Hint:

Question 15.
The angle of elevation of a cloud from a point h metres above a lake is β. The angle of depression of its reflection in the lake is 45°. The height of location of the cloud from the lake is

Solution:
(1) \(\frac{h(1+\tan \beta)}{1-\tan \beta}\)
Hint:

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Unit Exercise 4

Question 1.
In the figure, if BD⊥AC and CE ∠ AB, prove that

Solution:
In the figure’s ∆AEC and ∆ADB.
We have ∠AEC =∠ADB = 90 (∵ CE ∠AB and BD ∠AC)
and ∠EAC =∠DAB
[Each equal to ∠A]
Therefore by AA-criterion of similarity, we have ∆AEC ~ ∆ADB
(ii) We have
∆AEC ~ ∆ADB [As proved above]
⇒ \(\frac{C A}{B A}=\frac{E C}{D B} \Rightarrow \frac{C A}{A B}=\frac{C E}{D B}\)
Hence proved.

Question 2.
In the given figure AB||CD || EF . If AB = 6 cm, CD = x cm, EF = 4 cm, BD = 5 cm and DE = y cm. Find x and y.

Solution:
In the given figure, ∆AEF, and ∆ACD are similar ∆^s.
∠AEF = ∠ACD = 90°
∠A = ∠A (common)
∴ ∆AEF ~ ∆ACD (By AA criterion of similarity)

Subtstituting x = 2.4 cm in (3)

Question 3.
O is any point inside a triangle ABC. The bisector of ∠AOB , ∠BOC and ∠COA meet the sides AB, BC and CA in point D, E and F respectively.
Show that AD × BE × CF = DB × EC × FA
Solution:
In ∆AOB, OD is the bisector of ∠AOB.

In ∆BOC, OE is the bisector of ∠BOC
∴ \(\frac{\mathrm{OB}}{\mathrm{OC}}=\frac{\mathrm{BE}}{\mathrm{EC}}\) …………. (2)
In ∆COA, OF is the bisector of ∠COA.
∴ \(\frac{\mathrm{OC}}{\mathrm{OA}}=\frac{\mathrm{CF}}{\mathrm{FA}}\) …………… (3)
Multiplying the corresponding sides of (1), (2) and (3), we get

⇒ DB × EC × FA = AD × BE × CF
Hence proved.

Question 4.
In the figure, ABC is a triangle in which AB = AC . Points D and E are points on the side AB and AC respectively such that AD = AE . Show that the points B, C, E and D lie on a same circle.

Solution:
In order to prove that the points B, C, E and D are concyclic, it is sufficient to show that ∠ABC + ∠CED = 180° and ∠ACB + ∠BDE = 180°.

In ∆ABC, we have AB = AC and AD = AE.
⇒ AB – AD = AC – AE
⇒ DB = EC
Thus we have AD = AE and DB = EC. (By the converse of Thale’s theorem)
⇒ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) ⇒ DE||BC
∠ABC = ∠ADE (corresponding angles)
⇒ ∠ABC + ∠BDE = ∠ADE + ∠BDE (Adding ∠BDE on both sides)
⇒ ∠ABC + ∠BDE = 180°
⇒ ∠ACB + ∠BDE = 180° (∵ AB = AC ∴ ∠ABC = ∠ACB)
Again DE || BC
⇒ ∠ACB = ∠AED
⇒ ∠ACB + ∠CED = ∠AED + ∠CED (Adding ∠CED on both sides).
⇒ ∠ACB + ∠CED = 180° and
⇒ ∠ABC + ∠CED = 180° (∵ ∠ABC = ∠ACB)
Thus BDEC is a quadrilateral such that
⇒ ∠ACB + ∠BDE = 180° and
⇒ ∠ABC + ∠CED = 180°
∴ BDEC is a cyclic quadrilateral. Hence B, C, E, and D are concyclic points.

Question 5.
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels at a speed of 20 km/hr and the second train travels at 30 km/hr. After 2 hours, what is the distance between them?
Solution:
After 2 hours, let us assume that the first train is at A and the second is at B.

Question 6.
D is the mid point of side BC and AE⊥BC. If i BC a, AC = b, AB = c, ED = x, AD = p and AE = h , prove that
(i) b² = p² + ax + \(\frac{a^{2}}{4}\)
(ii) c² = p² – ax + \(\frac{a^{2}}{4}\)
(iii) b² + c² = 2p² + \(\frac{a^{2}}{2}\)
Solution:
From the figure, D is the mid point of BC.

We have ∠AED = 90°
∴ ∠ADE < 90° and ∠ADC > 90°
i.e. ∠ADE is acute and ∠ADC is obtuse,

(i) In ∆ADC, ∠ADC is obtuse angle.
AC² = AD² + DC² + 2DC × DE
⇒ AC² = AD² + \(\frac{1}{2}\) BC² + 2 . \(\frac{1}{2}\) BC . DE
⇒ AC² = AD² + \(\frac{1}{4}\) BC² + BC . DE
⇒ AC² = AD² + BC . DE + \(\frac{1}{4}\) BC²
⇒ b² = p² + ax + \(\frac{1}{4}\) a²
Hence proved.

(ii) In ∆ABD, ∠ADE is an acute angle.
AB² = AD² + BD² – 2BD . DE
⇒ AB² = AD² + \(\left(\frac{1}{2} \mathrm{BC}\right)^{2}\) – 2 × \(\frac{1}{2}\) BC . DE
⇒ AB² = AD² + \(\frac{1}{4}\) BC² – BC . DE
⇒ AB² = AD² – BC . DE + \(\frac{1}{4}\) BC²
⇒ c² = p² – ax + \(\frac{1}{4}\) a²
Hence proved.

(iii) From (i) and (ii) we get .
AB² + AC² = 2AD² + \(\frac{1}{2}\) BC²
i.e. c² + b² = 2p² + \(\frac{a^{2}}{2}\)
Hence it is proved.

Question 7.
A man whose eye-level is 2 m above the ground wishes to find the height of a tree. He places a mirror horizontally on the ground 20 m from the tree and finds that if he stands at a point C which is 4 m from the mirror B, he can see the reflection of the top of the tree. How height is the tree?
Solution:
From the figure; ∆DAC, ∆FBC are similar triangles and ∆ACE & ∆ABF are similar triangles.

∴ height of the tree h = 6x = 10 m.

Question 8.
An emu which is 8 ft tall standing at the foot of a pillar which is 30 ft height. It walks away from the pillar. The shadow of the emu falls beyond emu. What is the relation between the length of the shadow and the distance from the emu to the pillar?
Solution:
Let OA (emu shadow) the x and AB = y.
⇒ pillar’s shadow = OB = OA + AB
⇒ OB = x + y

Question 9.
Two circles intersect at A and B. From a point P on one of the circles lines PAC and PBD are drawn intersecting the second circle at C and D. Prove that CD is parallel to the tangent at P.
Solution:
Let XY be the tangent at P.
TPT: CD is || to XY.
Construction: Join AB.
ABCD is a cyclic quadilateral.
∠BAC + ∠BDC= 180° ………… (1)
∠BDC = 180° – ∠BAC …………. (2)
Equating (1) and (2)
we get ∠BDC = ∠PAB
Similarly we get ∠PBA = ∠ACD
as XY is tangent to the circle at ‘P’
∠BPY = ∠PAB (by alternate segment there)

∴ ∠PAB = ∠PDC
∠BPY = ∠PDC
XY is parallel of CD.
Hence proved.

Question 10.
Let ABC be a triangle and D, E, F are points on the respective sides AB, BC, AC (or their extensions). Let AD: DB = 5 : 3, BE : EC = 3 : 2 and AC = 21. Find the length of the line segment CF.
Solution: