Category: Class 10

Expert Teachers at SamacheerKalvi.Guru has created Tamil Nadu State Board New Syllabus Samacheer Kalvi 10th Social Science Model Question Papers 2019-2020 Pdf Free Download in English Medium and Tamil Medium of TN SSLC Class 10th Social Science Model Question Papers, Previous Year Question Papers, Sample Papers are part of Samacheer Kalvi 10th Model Question Papers 2020 Tamilnadu.

Here we have given the Government of TN State Board Samacheer Kalvi 10th Std Social Science Model Question Papers with Answers 2019-20 Tamil Medium Pdf. Students can view or download the Samacheer Kalvi Class 10th Social Science New Model Question Papers 2020 Tamil Nadu English Medium Pdf for their upcoming TN SSLC board examinations. Students can also read Tamilnadu Samcheer Kalvi 10th Social Science Book Solutions.

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• Samacheer Kalvi 10th Social Science Model Question Paper 1 
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• Samacheer Kalvi 10th Social Science Model Question Paper 1
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Samacheer Kalvi 10th Social Science Model Question Paper Design 2019-2020 Tamil Nadu

Samacheer Kalvi 10th Social Science Model Question Paper Weightage of Marks

It is necessary that students will understand the new pattern and style of Tamil Nadu Samacheer Kalvi 10th Social Science Model Question Papers 2020 according to the latest exam pattern. These State Board 10th Standard Social Science Public Exam Model Question Papers 2019-20 Tamil Nadu in Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN SSLC Board Exams and Score More marks.

We hope the given Tamil Nadu State Board New Syllabus Samacheer Kalvi Class 10th Social Science Model Question Papers 2019 2020 Pdf Free Download in English Medium and Tamil Medium will help you.

If you have any queries regarding the Government of TN State Board Samacheer Kalvi SSLC 10th Standard Social Science Model Question Papers with Answers 2019 20, Sample Papers, Previous Year Question Papers, drop a comment below and we will get back to you at the earliest.

Expert Teachers at SamacheerKalvi.Guru has created Tamil Nadu State Board New Syllabus Samacheer Kalvi 10th Science Model Question Papers 2019-2020 Pdf Free Download in English Medium and Tamil Medium of TN SSLC Class 10th Science Model Question Papers, Previous Year Question Papers, Sample Papers are part of Samacheer Kalvi 10th Model Question Papers 2020 Tamilnadu.

Here we have given the Government of TN State Board Samacheer Kalvi 10th Std Science Model Question Papers with Answers 2019-20 Tamil Medium Pdf. Students can view or download the Samacheer Kalvi Class 10th Science New Model Question Papers 2020 Tamil Nadu English Medium Pdf for their upcoming TN SSLC board examinations. Students can also read Tamilnadu Samcheer Kalvi 10th Science Book Solutions.

Tamil Nadu Samacheer Kalvi 10th Science Model Question Papers 2019 2020 Tamil English Medium

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• Samacheer Kalvi 10th Science Model Question Paper 1
• Samacheer Kalvi 10th Science Model Question Paper 2
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• Samacheer Kalvi 10th Science Model Question Paper 5 (Board Model Paper)

Tamil Nadu Samacheer Kalvi 10th Science Model Question Papers Tamil Medium

• Samacheer Kalvi 10th Science Model Question Paper 1
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Samacheer Kalvi 10th Science Model Question Paper Design 2019-2020 Tamil Nadu

Samacheer Kalvi 10th Science Model Question Paper Weightage of Marks

It is necessary that students will understand the new pattern and style of Tamil Nadu Samacheer Kalvi 10th Science Model Question Papers 2020 according to the latest exam pattern. These State Board 10th Standard Science Public Exam Model Question Papers 2019-20 Tamil Nadu in Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN SSLC Board Exams and Score More marks.

We hope the given Tamil Nadu State Board New Syllabus Samacheer Kalvi Class 10th Science Model Question Papers 2019 2020 Pdf Free Download in English Medium and Tamil Medium will help you.

If you have any queries regarding the Government of TN State Board Samacheer Kalvi SSLC 10th Standard Science Model Question Papers with Answers 2019 20, Sample Papers, Previous Year Question Papers, drop a comment below and we will get back to you at the earliest.

Expert Teachers at SamacheerKalvi.Guru has created Tamil Nadu State Board New Syllabus Samacheer Kalvi 10th Model Question Papers 2020-2021 Pdf Free Download in English Medium and Tamil Medium of TN SSLC Class 10th Model Question Papers, 10th Quarterly Half Yearly Model Question Papers Tamil Nadu, Previous Year Question Papers, Sample Papers, are part of Samacheer Kalvi.

Here we have given the Government of TN State Board Samacheer Kalvi 10th Std Model Question Papers for All Subjects with Answers 2020-21 Tamil Medium Pdf. Students can view or download the Samacheer Kalvi Class 10th New Model Question Papers 2021 Tamil Nadu English Medium Pdf for their upcoming TN SSLC board examinations. Students can also read Tamilnadu Samcheer Kalvi 10th Books Solutions.

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Samacheer Kalvi 10th Model Question Papers for All Subjects

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It is necessary that students will understand the new pattern and style of Tamil Nadu Board Model Question Papers for 10th Standard Samacheer Kalvi according to the latest exam pattern. These State Board 10th Public Model Question Papers 2020-21 Tamil Nadu in Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN SSLC Board Exams and Score More marks.

We hope the given Tamil Nadu State Board New Syllabus Samacheer Kalvi Class 10th Model Question Papers 2020 2021 Pdf Free Download in English Medium and Tamil Medium will help you.

If you have any queries regarding the Government of TN State Board Samacheer Kalvi SSLC 10th Standard Model Question Papers with Answers 2020 21, 10th Quarterly Half Yearly Model Question Papers Tamil Nadu, Sample Papers, Previous Year Question Papers, drop a comment below and we will get back to you at the earliest.

Expert Teachers at SamacheerKalvi.Guru has created Tamil Nadu State Board New Syllabus Samacheer Kalvi 10th Maths Model Question Papers 2019-2020 Pdf Free Download in English Medium and Tamil Medium of TN SSLC Class 10th Maths Model Question Papers, Previous Year Question Papers, Sample Papers are part of Samacheer Kalvi 10th Model Question Papers 2020 Tamilnadu.

Here we have given the Government of TN State Board Samacheer Kalvi 10th Std Maths Model Question Papers with Answers 2019-20 Tamil Medium Pdf. Students can view or download the Samacheer Kalvi Class 10th Maths New Model Question Papers 2020 Tamil Nadu English Medium Pdf for their upcoming TN SSLC board examinations. Students can also read Tamilnadu Samcheer Kalvi 10th Maths Book Solutions.

Tamil Nadu Samacheer Kalvi 10th Maths Model Question Papers 2019 2020 Tamil English Medium

Tamil Nadu Samacheer Kalvi 10th Maths Model Question Papers English Medium

• Samacheer Kalvi 10th Maths Model Question Paper 1
• Samacheer Kalvi 10th Maths Model Question Paper 2
• Samacheer Kalvi 10th Maths Model Question Paper 3
• Samacheer Kalvi 10th Maths Model Question Paper 4
• Samacheer Kalvi 10th Maths Model Question Paper 5 (Board Model Paper)

Tamil Nadu Samacheer Kalvi 10th Maths Model Question Papers Tamil Medium

• Samacheer Kalvi 10th Maths Model Question Paper 1
• Samacheer Kalvi 10th Maths Model Question Paper 2
• Samacheer Kalvi 10th Maths Model Question Paper 3
• Samacheer Kalvi 10th Maths Model Question Paper 4
• Samacheer Kalvi 10th Maths Model Question Paper 5

Samacheer Kalvi 10th Maths Model Question Paper Design 2019-2020 Tamil Nadu

Samacheer Kalvi 10th Maths Model Question Paper Weightage of Marks

It is necessary that students will understand the new pattern and style of Tamil Nadu Samacheer Kalvi 10th Maths Model Question Papers 2020 according to the latest exam pattern. These State Board 10th Standard Maths Public Exam Model Question Papers 2019-20 Tamil Nadu in Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN SSLC Board Exams and Score More marks.

We hope the given Tamil Nadu State Board New Syllabus Samacheer Kalvi Class 10th Maths Model Question Papers 2019 2020 Pdf Free Download in English Medium and Tamil Medium will help you.

If you have any queries regarding the Government of TN State Board Samacheer Kalvi SSLC 10th Standard Maths Model Question Papers with Answers 2019 20, Sample Papers, Previous Year Question Papers, drop a comment below and we will get back to you at the earliest.

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: