Category: Class 9

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.3

Question 1.
The diameter of the circle is 52 cm and the length of one of its chord is 20 cm. Find the distance of the chord from the centre.
Radius of a circle = \(\frac{56}{2}\) = 26 cm
Solution:

Length of the chord = 20 cm
AC = \(\frac{20}{2}\)
= 10 cm
In ΔOAC, OC² = OA² – AC²
= 26² – 10²
= (26 + 10) (26 – 10)
= 36 × 16
OC = \(\sqrt{30×16}\)
= 6 × 4 cm
= 24 cm
Distance of the chord from the centre = 24 cm.

Question 2.
The chord of length 30 cm is drawn at the distance of 8 cm from the centre of the circle. Find the radius of the circle.
Solution:

Distance AC = \(\frac{1}{2}\) × Length of chord
= \(\frac{1}{2}\) × 30
= 15 cm
Distance from the centre = 8 cm
In ΔOAC Radius (OA) = \(\sqrt{AC² + OC²}\)
= \(\sqrt{15² + 8²}\)
= \(\sqrt{224 + 64}\)
= \(\sqrt{289}\)
= 17
Radius of the circle = 17 cm.

Question 3.
Find the length of the chord AC where AB and CD are the two diameters perpendicular to each other of a circle with radius 4√2 cm and also find ∠OAC and ∠OCA.
Solution:
Radius of a circle = 4√2 cm
In the right ΔAOC,

AC² = OA² + OC²
AC² = (4√2)² + (4√2)²
= 32 + 32 = 64
AC = \(\sqrt{64}\)
= 8
Length of the chord = 8 cm,
∠OAC = ∠OCA = 45°
Since OAC is an isosceles right angle triangle.

Question 4.
A chord is 12 cm away from the centre of the circle of radius 15cm. Find the length of the chord.
Solution:
Radius of a circle (OA) = 15 cm
Distance from centre to the chord (OC) = 12 cm
In the right ΔOAC,

AC² = OA² – OC²
= 15² – 12²
= 225 – 144
= 81
AC = \(\sqrt{81}\)
= 9
Length of the chord (AB) = AC + CB
= 9 + 9 = 18 cm.

Question 5.
In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such that AB = 16 cm and CD = 12 cm determine the distance between the two chords?
Solution:
Length of the chord (AB) = 16 cm
∴ AF = \(\frac{1}{2}\) × 16
= 8 cm
Length of the chord (CD) = 12 cm
∴ CE = \(\frac{1}{2}\) × 12
= 6 cm

In the right ΔOCE,
OE² = OC² – CE²
= 10² – 6²
= 100 – 36
= 64
OE = \(\sqrt{64}\)
= 8 cm
In the right ΔOAF,
OF² = OA² – AF²
= 10² – 8²
= 100 – 64
= 36
OE = \(\sqrt{36}\)
= 6 cm
Distance between the two chords = OE + OF
= 8 + 6
= 14 cm.

Question 6.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Ans

Two circules intersect at C and D
CD is the common chord.
CD = AC + AD
= 3 + 3
= 6 cm
Length of the common chord = 6 cm.

Question 7.
Find the value of x° in the following figures:

Solution:
(i) ∠BOC = 30° + 60°
= 90°
∠BAC (x) = \(\frac{1}{2}\) ∠BOC (By theorem)
= \(\frac{1}{2}\) × 90°
x = 45°

(ii) Join OP
∠QPR = \(\frac{80}{2}\) = 40° (Angle of the circumference is \(\frac{1}{2}\) the angle at the centre)
∠RPO = 30° (OP and OR are equal sides)
∠OPQ = 40° – 30°
= 10°
∠OQP = 10° (OQ and OP are equal sides)
∴ x = 10°

(iii) ∠OPN = ∠ONP (ON and OP are the radius of the circle)
= 90°
In ΔOPN
∠PON + ∠ONP + ∠NPO = 180° (Sum of the angles of a triangle)
70° + x° + x° = 180°
2x° + 70° = 180°
2x = 180° – 70°
2x = 110°
X = \(\frac{110°}{2}\)
= 55°
The value of x = 55°

(iv) Reflex ∠YOZ = 2∠YXZ
= 2(120°)
= 240°
∠YOZ = 360° – reflex ∠YOZ
= 360° – 240°
= 120°
∴ x = 120°

(v) ∠OAC + ∠OCA = 180° – 100°
= 80°
∠OAC = \(\frac{1}{2}\) × 80° [Since OA = OC, (∴ ∠OAC = ∠OCA)]
= 40°
∠OBA + ∠OAB = 180° – 140°
= 40° [Since ∠OBA = ∠OAB, since OB = OA]
∴∠OAB = \(\frac{40°}{2}\)
= 20°
∠BAC = ∠OAB + ∠OAC
= 20° + 40°
x = 60°

Question 8.
In the given figure, ∠CAB = 25°, find ∠BDC, ∠DBA and ∠COB.

Solution:
In ΔACP,
∠ACP = 180° – (25° + 90°)
= 180° – 115°
= 65°
∠CBA = ∠CAB = 25° [Both the angles are standing in the same base]
∠DBA = 65° [∠DBA and ∠BCA standing in the same base]
∠COB = 50°

Students can download Maths Chapter 5 Coordinate Geometry Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions

I. Multiple Choice Questions.

Question 1.
On which quadrant does the point (- 4, 3) lie?
(a) I
(b) II
(c) III
(d) IV
Solution:
(b) II

Question 2.
The point whose abscissa is 5 and lies on the x-axis is …….
(a) (-5, 0)
(b) (5, 5)
(c) (0, 5)
(d) (5, 0)
Solution:
(d) (5, 0)

Question 3.
A point which lies in the III quadrant is ……..
(a) (5, 4)
(b) (5, -4)
(c) (-5, -4)
(d) (-5, 4)
Solution:
(c) (-5, -4)

Question 4.
A point on the y-axis is ……..
(a) (1, 1)
(b) (6, 0)
(c) (0, 6)
(d) (-1, -1)
Solution:
(c) (0, 6)

Question 5.
The distance between the points (4, -1) and the origin is ……..
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(d) \(\sqrt{17}\)

Question 6.
The distance between the points (-1, 2) and (3, 2) is ……..
(a) \(\sqrt{14}\)
(b) \(\sqrt{15}\)
(c) 4
(d) 0
Solution:
(c) 4

Question 7.
The centre of a circle is (0, 0). One end point of a diameter is (5, -1), then the radius is …….
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(c) \(\sqrt{26}\)

Question 8.
The point (0, -3) lies on
(a) + ve x-axis
(b) + ve y-axis
(c) – ve x-axis
(d) – ve y-axis
Solution:
(d) – ve y-axis

Question 9.
The point which is on y-axis with ordinate -5 is ……..
(a) (0, -5)
(b) (-5, 0)
(c) (5, 0)
(d) (0, 5)
Solution:
(a) (0, -5)

Question 10.
The diagonal of a square formed by the points (1, 0), (0, 1), (-1, 0) and (0, -1) is …….
(a) 2
(b) 4
(c) √2
(d) 8
Solution:
(a) 2

Question 11.
The distance between the points (-2, 2) and (3, 2) is ……..
(a) 10 units
(b) 5 units
(c) 5√3 units
(d) 20 units
Solution:
(b) 5 units

Question 12.
The midpoint of the line joining the points (1, -1) and (-5, 3) is ……..
(a) (2, 1)
(b) (2, -1)
(c) (-2, -1)
(d) (-2, 1)
Solution:
(d) (-2, 1)

Question 13.
If the centroid of a triangle is at (1, 3) and two of its vertices are (-7, 6) and (8, 5) then the third vertex is ……..
(a) (-2, 2)
(b) (2, -2)
(c) (-2, -2)
(d) (2, 2)
Solution:
(b) (2, -2)

Question 14.
The ratio in which the X-axis divides the line segment joining the points (6, 4) and (1, -7) is ……..
(a) 1 : 2
(b) 2 : 3
(c) 4 : 7
(d) 7 : 4
Solution:
(c) 4 : 7

Question 15.
The centroid of a triangle (3, -5), (-7, 4) and (10, -2) is …….
(a) (2, -1)
(b) (2, 1)
(c) (-2, 1)
(d) (1, -2)
Solution:
(a) (2, -1)

II. Answer the Following Questions.

Question 1.
Show that the given points (1, 1), (5, 4), (-2, 5) are the vertices of an isosceles right angled triangle.
Solution:
Let A (1, 1), B (5, 4) and G (-2, 5)

AB = 5, AC = 5
∴ ABC is an isosceles triangle …….. (1)
BC² = AB² + AC²
50 = 25 + 25 ⇒ 50 = 50
∴ ∠A = 90° ……… (2)
From (1) and (2) we get ABC is an isosceles right angle triangle.

Question 2.
Show that the point (3, -2), (3, 2), (-1, 2) and (-1, -2) taken in order are the vertices of a square.
Solution:

= \(\sqrt{16}\)
= 4
AB = BC = CD = DA = 4. All the four sides are equal.
∴ ABCD is a Rhombus ……..(1)

Diagonal AC = Diagonal BD = \(\sqrt{32}\) ……..(2)
From (1) and (2) we get ABCD is a square.

Question 3.
Show that the point A (3, 7) B (6, 5) and C (15, -1) are collinear.
Solution:

AB + BC = AC ⇒ \(\sqrt{13}\) + 3\(\sqrt{13}\) = 4\(\sqrt{13}\)
∴ The points A, B, C are collinear.

Question 4.
Find the type of triangle formed by (-1, -1), (1, 1) and (-√, √3)
Solution:
Let the point A (-1, -1), B (1, 1) and C (-√3, √3)

AB = BC = AC = √8
∴ ABC is an equilateral triangle.

Question 5.
Find x such that PQ = QR where P(6, -1) Q(1, 3) and R(x, 8) respectively.
Solution:

But PQ = QR
\(\sqrt{(x-1)^{2}+25}\) = \(\sqrt{41}\)
Squaring on both sides
(x – 1)² + 25 = 41
(x – 1)² = 41 – 25 = 16
x – 1 = \(\sqrt{16}\) = ± 4
x – 1 = 4 (or) x – 1 = – 4
x = 5 (or) x = -4 + 1 = -3
The value of x = 5 or – 3

Question 6.
Find the coordinate of the point of trisection of the line segment joining (4, -1) and
Solution:
Let A (4, -1) and B (-2, -3) are the given points
Let P (a, b) and Q (c, d) be the points of trisection of AB.
∴ AP = PQ = QB

The required coordinate P is (2, –\(\frac{5}{3}\)) and Q is (0, –\(\frac{7}{3}\))

Question 7.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Given points are A(-3, 10), B(6, -8) and P(-1, 6)
divides AB internally in the ratio m : n
By section formula.
A line divides internally in the ratio m : n the point P =

∴ \(\frac{6m-3n}{m+n}\) = -1
6m – 3n = -m – n
6m + m = 3n – n
7m = 2n ⇒ \(\frac{m}{n}\) = \(\frac{2}{7}\)
∴ m : n = 2 : 7
and
\(\frac{-8m+10n}{m+n}\) = 6
-8m + 10n = 6m + 6n
-8m – 6m = 6n – 10n
14m = 4n
∴ \(\frac{m}{n}\) = \(\frac{14}{4}\) = \(\frac{2}{7}\)
Hence P divides AB internally in the ratio 2 : 7

Question 8.
If (1, 2) (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find “x” and “y”.
Solution:
Let A(1, 2), B(4, y), C(x, 6) and D(3, 5)

Since ABCD is a parallelogram the diagonal bisect each other
Mid point of AC = Mid point of BD
(\(\frac{1+x}{2}\), 4) = (\(\frac{7}{2}\), \(\frac{y+5}{2}\))
\(\frac{1+x}{2}\) = \(\frac{7}{2}\)
1 + x = 7
x = 7 – 1
= 6
and
\(\frac{y+5}{2}\) = 4
y + 5 = 8
y = 8 – 5
= 3
∴ The value of x = 6 and y = 3

Students can download Maths Chapter 6 Trigonometry Ex 6.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.3

Question 1.
Find the value of the following:

(iii) tan 15° tan 30° tan 45° tan 60° tan 75°

Solution:
(i) cos 45° = \(\frac{1}{√2}\)

= 1² + 1² – 2(\(\frac{1}{√2}\))²
= 1 + 1 – 2(\(\frac{1}{2}\))
= 2 – 1
= 1

(ii) cos 60° = \(\frac{1}{2}\)

= 1 + 1 + 1 – 8(\(\frac{1}{2}\))²
= 3 – 8 × \(\frac{1}{4}\)
= 3 – 2
= 1

(iii) tan 30° = \(\frac{1}{√3}\), tan 45° = 1, tan 60° = √3
tan 15° . tan 30°. tan 45° . tan 60°. tan 75° = tan 15° . \(\frac{1}{√3}\) . 1 . √3 tan 75°
= tan 15° × tan 75° × \(\frac{1}{√3}\) × 1 × √3
= tan(90° – 75°) × \(\frac{1}{cot 75°}\) × 1 [tan 90° – θ = cot θ]
= cot 75° × \(\frac{1}{cot 75°}\) × 1
= 1

= 1 + 1
= 2

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.2

Question 1.
Verify the following equalities:
(i) sin² 60° + cos² 60° = 1
Solution:
sin 60° = \(\frac{√3}{2}\); cos 60° = \(\frac{1}{2}\)
L.H.S = sin² 60° + cos² 60°

L.H.S = R.H.S
Hence it is proved.

(ii) 1 + tan² 30° = sec² 30°
Solution:
tan 30° = \(\frac{1}{√3}\); sec 30° = \(\frac{2}{√3}\)
L.H.S = 1 + tan² 30°

∴ L.H.S = R.H.S
Hence it is proved.

(iii) cos 90° = 1 – 2sin² 45° = 2cos² 45° – 1
Solution:
cos 90° = 0, sin 45° = \(\frac{1}{√2}\), cos 45° = \(\frac{1}{√2}\)
cos 90° = 0 ……. (1)
1 – 2 sin² 45° = 1 – 2 (\(\frac{1}{√2}\))²
= 1 – 2 × \(\frac{1}{2}\)
= 1 – 1 = 0 → (2)
2 cos² 45° – 1 = 2(\(\frac{1}{√2}\))² – 1
= \(\frac{2}{2}\) – 1
= \(\frac{2 – 2}{2}\) = 0 → (3)
From (1), (2) and (3) we get
cos 90° = 1 – 2 sin² 45° = 2 cos² 45° – 1
Hence it is proved.

(iv) sin 30° cos 60° + cos 30° sin 60° = sin 90°
Solution:
sin 30° = \(\frac{1}{2}\); cos 60° = \(\frac{1}{2}\); cos 30° = \(\frac{√3}{2}\); sin 60° = \(\frac{√3}{2}\); sin 90° = 1
L.H.S = sin 30° cos 60° + cos 30° sin 60°

= 1
R.H.S = sin 90° = 1
L.H.S = R.H.S
Hence it is proved.

Question 2.
Find the value of the following:
(i) \(\frac{tan 45°}{cosec 30°}\) + \(\frac{sec 60°}{cot 45°}\) – \(\frac{5 sin 90°}{2 cos 0°}\)
(ii) (sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)
(iii) sin²30° – 2 cos³ 60° + 3 tan⁴ 45°
Solution:
(i) tan 45° = 1, cosec 30° = 2; sec 60° = 2; cot 45° = 1; tan 45°, sin 90° = 1; cos 0° = 1

= 0

(ii) sin 90° = 1; cos 60° = \(\frac{1}{2}\); cos 45° = \(\frac{1}{√2}\); sin 30° = \(\frac{1}{2}\); cos 0° = 1
(sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)

= \(\frac{7}{4}\)

(iii) sin 30° = \(\frac{1}{2}\); cos 60° = \(\frac{1}{2}\); tan 45° = 1
sin² 30° – 2 cos³ 60° + 3 tan⁴ 45°

= 3

Question 3.
Verify cos 3 A = 4 cos³ A – 3 cos A, when A = 30°
Solution:
L.H.S = cos 3 A
= cos 3 (30°)
= cos 90°
= 0
R.H.S = 4 cos³ A – 3 cos A
= 4 cos³ 30° – 3 cos 30°

= 0
∴ L.H.S = R.H.S
Hence it is proved.

Question 4.
Find the value of 8 sin2x, cos 4x, sin 6x, when x = 15°.
Solution:
8 sin 2x cos 4x sin 6x = 8 sin 2 (15°) × cos 4 (15°) × sin (6 × 15°)
= 8 sin 30° × cos 60° × sin 90°
= 8 × \(\frac{1}{2}\) × \(\frac{1}{2}\) × 1
= 2

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.5

Question 1.
Find the centroid of the triangle whose vertices are
(i) (2, -4), (-3, -7) and (7, 2)
Solution:
Let the vertices of a triangle be A (2, -4), B (-3, -7) and C (7, 2) Centroid

Centroid is (2, -3)

(ii) (-5, -5), (1, -4) and (-4, -2)
Solution:
Let the vertices of a triangle be A (-5, -5), B (1, -4) and C (-4, -2)

Question 2.
If the centroid of a triangle is at (4, -2) and two of its vertices are (3, -2) and (5, 2) then find the third vertex of the triangle.
Solution:
Let the vertices of a triangle be A (3, -2), B (5, 2) and C (x₃, y₃)
Centroid of a triangle is (4, -2)

∴ \(\frac{8+x_{3}}{3}\) = 4
8 + x₃ = 12
x₃ = 12 – 8
= 4
and
\(\frac{y_{3}}{3}\) = -2
y₃ = -6
∴ The third vertex is (4, -6)

Question 3.
Find the length of median through A of a triangle whose vertices are A(-1, 3), B (1, -1) and C (5, 1).
Solution:
AD is the median of the ΔABC
D is the mid-point of BC

Length of the median AD is 5 units.

Question 4.
The vertices of a triangle are (1, 2), (h, -3) and (-4, k). If the centroid of the triangle is at the point (5, -1) then find the value of \(\sqrt{(h+k)^{2}+(h+3k)^{2}}\)
Solution:
Let the vertices A (1, 2), B (h, -3) and C (-4, k)


\(\frac{-3+h}{3}\) = 5
-3 + h = 15
h = 15 + 3 = 18
and
\(\frac{-1+k}{3}\) = -1
-1 + k = -3
k = -3 + 1
k = -2

Question 5.
Orthocentre and centroid of a triangle are A(-3, 5) and B(3, 3) respectively. If C is the circumcentre and AC is the diameter of this circle, then find the radius of the circle.
Solution:
Let PQR be any triangle orthocentre, centroid and circumcentre.

A orthocentre is (-3, 5)
B centroid is (3, 3)
C orthocentre is (a, 6)
Also \(\frac{AB}{BC}\) = \(\frac{2}{1}\)
B divides AC in the ratio 2 : 1
A line divides internally in the ratio point P is (\(\frac{mx_{2}+nx_{1}}{m+n}\), \(\frac{my_{2}+ny_{1}}{m+n}\))
m = 2
x₁ = 3
y₁ = 5
amd
n = 1
x₂ = a
y₂ = b
∴ The point B (\(\frac{2a-3}{2+1}\), \(\frac{2b+5}{2+1}\))
(3, 3) = (\(\frac{2a-3}{3}\), \(\frac{2b+5}{3}\))
\(\frac{2a-3}{3}\) = 3
2a – 3 = 9
2a = 9 + 3
2a = 12
a = \(\frac{12}{2}\) = 6
and
\(\frac{2b+5}{3}\)
2b + 5 = 9
2b = 9 – 5
2b = 4
b = \(\frac{4}{2}\) = 2
∴ Orthocentre C is (6, 2)

Question 6.
ABC is a triangle whose vertices are A (3, 4), B (-2, -1) and C (5, 3). If G is the centroid and BDCG is a parallelogram then find the coordinates of the vertex D.
Solution:
The vertices of a triangle are A (3, 4), B (-2, -1) and C (5, 3)

The point G is (2, 2)
Let the vertices D be (a, b)
Since BDCG is a parallelogram
Mid-point of BC = Mid-point of DG

\(\frac{2+a}{2}\) = \(\frac{3}{2}\)
4 + 2a = 6
2a = 6 – 4
2a = 2
a = 1
and
\(\frac{2+b}{2}\) = 1
2 + b = 2
b = 2 – 2 = 0
The vertices D is (1, 0).

Question 7.

Solution:
In ΔABC, Let A (x₁, y₁), B (x₂, y₁) and C (x₃, y₃)

x₁ + x₂ = 3 → (1)
y₁ + y₂ = 10 → (2)

x₂ + x₃ = 14 → (3)
y₂ + y₃ = 10 → (4)

x₁ + x₃ = 3 → (5)
y₁ + y₃ = 10 → (6)
Add (1) + (3) + (5) We get
2x₁ + 2x₂ + 2x₃ = 30
2(x₁ + x₂ + x₃) = 30
x₁ + x₂ + x₃ = 15
From (1), x₁ + x₂ = 3
∴ x₃ = 12
From (3), x₂ + x₃ = 14
∴ x₁ = 1
From (5), x₁ + x₃ = 13
∴ x₂ = 2
Add (2), (4) and (6) we get
2y₁ + 2y₂ + 2y₃ = -12
2(y₁ + y₂ + y₃) = – 12
∴ y₁ + y₂ + y₃ = -6
From (2), y₁ + y₂ = 10
∴ y₃ = -16
From (4) y₂ + y₃ = -9
∴ y₁ = 3
From (6) y₁ + y₃ = -13
∴ y₂ = 7
The vertices of the A are A (1, 3), B (2, 7) and C (12, -16)

The Centroid of Δ is (5, -2)

Students can download Maths Chapter 6 Trigonometry Ex 6.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.1

Question 1.
From the given figure, find all the trigonometric ratios of angle B.

Solution:

Question 2.
From the given figure, find the values of

(i) sin B
(ii) sec B
(iii) cot B
(v) tan C
(vi) cosec C
Solution:
In the right ΔABD,
AD² = AB² – BD²
= 13² – 5²
= 169 – 25
= 144
AD = \(\sqrt{144}\)
= 12
In the right ΔADC,
AC² = AD² + DC²
= 12² + 16²
= 144 + 256
= 400
AC = \(\sqrt{400}\)
= 20

Question 3.
If 2 cos θ = √3, then find all the trigonometric ratios of angle θ.
Solution:

2 cos θ = √3 ⇒ cos θ = \(\frac{√3}{2}\)
AB² = AC² – BC²
= 2² – (√3)² ⇒ = 4 – 3 = 1
AB = √1 = 1

Question 4.
If cos A =\(\frac{3}{5}\), then find the value of \(\frac{sin A-cos A}{2 tan A}\)
Solution:

cos A = \(\frac{3}{5}\)
ΔABC
BC² = AC² – AB²
= 5² – 3²
= 25 – 9
= 16
BC = \(\sqrt{16}\) = 4
sin A = \(\frac{4}{5}\); tan A = \(\frac{4}{3}\)

∴ The value of \(\frac{sin A-cos A}{2 tan A}\) = \(\frac{3}{40}\)

Question 5.
If cos A = \(\frac{2x}{1+x_{2}}\) then find the values of sin A and tan A in terms of x.
Solution:
cos A = \(\frac{2x}{1+x_{2}}\)
In the triangle ABC

BC² = AC² – AB²
= (1 + x²)² – (2x)²
= 1 + x⁴ + 2x² – 4x²
= x⁴ – 2x² + 1
= (x² – 1)² (or) (1 – x²)² (using (a – b)²)
BC = \(\sqrt{(x^{2}-1)^{2}}\) (or) \(\sqrt{(1-x^{2})^{2}}\)
BC = x² – 1
The value of sin A = \(\frac{BC}{AC}\) = \(\frac{x²-1}{x²+1}\)
tan A = \(\frac{BC}{AB}\) = \(\frac{x²-1}{2x}\)
and
BC = 1 – x²
The value of sin A = \(\frac{1-x²}{1+x²}\)
tan A = \(\frac{1-x²}{2x}\)

Question 6.
If sin θ = \(\frac{a}{\sqrt{a²+b²}}\) then show that b sin θ = a cos θ.
Solution:

sin θ = \(\frac{a}{\sqrt{a²+b²}}\)
In the triangle ΔABC
BC² = AC² – AB²

L. H. S = R. H. S

Question 7.
If 3 cot A = 2, then find the value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)
Solution:

3 cot A = 2 ⇒ cot A = \(\frac{2}{3}\)
AC² = AB² + BC²
= 3² + 2²
= 9 + 4
AC = \(\sqrt{13}\)
cos A = \(\frac{AB}{AC}\) = \(\frac{3}{\sqrt{13}}\)
sin A = \(\frac{BC}{AC}\) = \(\frac{2}{\sqrt{13}}\)
The value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)

The value is (\(\frac{-1}{13}\))

Question 8.
If cos θ : sin θ = 1 : 2, then find the value of \(\frac{8cos θ-2cos θ}{4 cos θ+2 sin θ}\)
Solution:
cos θ : sin θ = 1 : 2

Aliter:
cos θ : sin θ = 1 : 2
2 cos θ = sin θ ⇒ 2 = \(\frac{sin θ}{cos θ}\) ⇒ 2 = tan θ

∴ The value is \(\frac{1}{2}\)

Question 9.
From the given figure, prove that θ + ∅ = 90°. Also prove that there are two other right angled triangles. Find sin α, cos β and tan ∅.

Solution:
In the ΔABC,
AB = 9 + 16 = 25
AC = 15; BC = 20
AB² = 25²
= 625 ……. (1)
AC² + BC² = 15² + 20²
= 225 + 400
= 625 …….. (2)
From (1) and (2) we get
AB² = AC² + BC²
ABC is a right angle triangle at C (Pythagoras theorem)
∴ ∠C = 90°
θ + ∅ = 90°
Also ADC is a right angle triangle ∠ADC = 90° (Given)
BDC is also a right angle triangle ∠BDC = 90° (since ADB is a straight line sum of the two angle is 180°)
From the given diagram

Question 10.
A boy standing at a point O finds his kite flying at a point P with distance OP = 25 m. It is at a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground? (use trigonometric ratios).

Solution:
Let the angle O be “θ”
In ΔONQ

In ΔOMP
sin θ = \(\frac{PM}{OP}\) ⇒ sin θ = \(\frac{5}{25}\)
sin θ = \(\frac{1}{5}\) ……… (2)
From (1) and (2) we get
\(\frac{h}{35}\) = \(\frac{1}{5}\)
5 h = 35 ⇒ h = \(\frac{35}{5}\) = 7
The height of the kite from the ground is 7m.

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.5

Question 1.
If sin 30° = x and cos 60° = y, then x² + y² is …….
(a) \(\frac{1}{2}\)
(b) 0
(c) sin 90°
(d) cos 90°
Solution:
(a) \(\frac{1}{2}\)
Hint:
sin 30° = x = \(\frac{1}{2}\)
cos 60° = y = \(\frac{1}{2}\)
x² + y²

Question 2.
If tan θ = cot 37°, then the value of θ is ………
(a) 37°
(b) 53°
(c) 90°
(d) 1°
Solution:
(b) 53°
Hint:
tan θ = cot 37°
= cot (90° – 53°)
= tan 53°
The value of θ is 53°

Question 3.
The value of tan 72°. tan 18° is ………
(a) 0
(b) 1
(c) 18°
(d) 72°
Solution:
(b) 1
Hint:
tan 72° . tan 18° = tan 72° . tan (90° – 72°)
= tan 72° . cot 72°
= tan 72° × \(\frac{1}{tan 72°}\)
= 1

Question 4.
The value of \(\frac{2 tan 30°}{1 – tan^{2} 30°}\) is equal to ………
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°
Solution:
(c) tan 60°
Hint:

= √3 = tan 60°

Question 5.
If 2 sin 2θ = √3 then the value of θ is ………
(a) 90°
(b) 30°
(c) 45°
(d) 60°
Solution:
(b) 30°
Hint:
2 sin 2θ = √3 ⇒ sin 2θ = \(\frac{√3}{2}\)
sin 2θ = sin 60° ⇒ 2θ = 60°
θ = 30°

Question 6.
The value of 3 sin 70° sec 20° + 2 sin 39° sec 51° is ………
(a) 2
(b) 3
(c) 5
(d) 6
Solution:
(c) 5
Hint:
3 sin 70° sec 20° + 2 sin 39° sec 51°
= 3. sin 70° . sec (90° – 70°) + 2 sin 39° . sec (90° – 39°)
= 3. sin 70° . cosec 70° + 2 sin 39° . cosec 39°
= 3. sin70° × \(\frac{1}{sin 70°}\) + 2 sin 39° × \(\frac{1}{sin 39°}\)
= 3 + 2
= 5

Question 7.
The value of \(\frac{1-tan^{2}45°}{1+tan^{2}45°}\) is ……..
(a) 2
(b) 1
(c) 0
(d) \(\frac{1}{2}\)
Solution:
(c) 0
Hint:
\(\frac{1-tan^{2}45°}{1+tan^{2}45°}\) = \(\frac{1-1}{1+1}\)
= \(\frac{0}{2}\) = 0

Question 8.
The value of cosec (70° + θ) – sec (20° + θ) + tan (65° + θ) – cot (25° + θ) is ……..
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
cosec (70° + θ) – sec (20° – θ) + tan (65° + θ) – cot (25° – θ)
= sec [90° – (70° + θ)] – sec (20° – θ) + tan (65° + θ) – tan [90° – (25° – θ)]
= sec (20° – θ) – sec (20° – θ) + tan (65° + θ) – tan (65° + θ)
= 0

Question 9.
The value of tan 1° . tan 2° . tan 3°…. tan 89° is ………
(a) 0
(b) 1
(c) 2
(d) \(\frac{√3}{2}\)
Solution:
(b) 1
Hint:
tan 1° . tan 2° . tan 3° …….. tan 89°
= tan (90° – 89°). tan (90° – 88°) .tan (90° – 87°) …….. tan 45° . tan (89°)
= cot 89° . cot 88°. cot 87°. ……. tan 45° …….. tan 87°. tan 88°. tan 89°
= 1

Question 10.
Given that sin α = \(\frac{1}{2}\) and cos β = \(\frac{1}{2}\), then the value of α + β is ………
(a) 0°
(b) 90°
(c) 30°
(d) 60°
Solution:
Hint:
sin α = \(\frac{1}{2}\)
sin 30° = \(\frac{1}{2}\) ∴ α = 30°
cos β = \(\frac{1}{2}\) ∴ β = 60°
∴ α + β = 30° + 60°
= 90°

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.6 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.6

Question 1.
If the y-coordinate of a point is zero, then the point always lies ……..
(a) in the I quadrant
(b) in the II quadrant
(c) on x-axis
(d) on y-axis
Solution:
(c) on x-axis

Question 2.
The points (-5, 2) and (2, -5) lie in the ……….
(a) same quadrant
(b) II and III quadrant respectively
(c) II and IV quadrant respectively
(d) IV and II quadrant respectively
Solution:
(c) II and IV quadrant respectively

Question 3.
On plotting the points O (0, 0), A (3, -4), B (3, 4) and C (0, 4) and joining OA, AB, BC and CO, which of the following figure is obtained?
(a) Square
(b) Rectangle
(c) Trapezium
(d) Rhombus
Solution:
(c) Trapezium

Question 4.
If P (-1, 1), Q ( 3, -4), R (1, -1), S (-2, -3) and T (- 4, 4) are plotted on a graph paper, then the points in the fourth quadrant are ………
(a) P and T
(b) Q and R
(c) only S
(d) P and Q
Solution:
(b) Q and R

Question 5.
The point whose ordinate is 4 and which lies on the v-axis is ……….
(a) (4, 0)
(b) (0, 4)
(c) (1, 4)
(d) (4, 2)
Solution:
(b) (0, 4)

Question 6.
The distance between the two points (2, 3) and (1, 4) is ……..
(a) 2
(b) \(\sqrt{56}\)
(c) \(\sqrt{10}\)
(d) √2
Solution:
(d) √2
Hint:
\(\sqrt{(1-2)^{2}+(4+3)^{2}}\)
= \(\sqrt{(-1)^{2}+1^{2}}\)
= \(\sqrt{1+1}\)
= √2

Question 7.
If the points A (2, 0), B (-6, 0), C (3, a – 3) lie on the x-axis then the value of a is ……..
(a) 0
(b) 2
(c) 3
(d) -6
Solution:
(c) 3
Hint:
a – 3 = 0 ⇒ a = 3

Question 8.
If (x + 2, 4) = (5, y – 2), then the coordinates (x, y) are ………
(a) (7, 12)
(b) (6, 3)
(c) (3, 6)
(d) (2, 1)
Solution:
(c) (3, 6)
Hint:
x + 2 = 5
∴ x = 5 – 2 = 3
and
4 = y – 2
4 + 2 = y
∴ y = 6

Question 9.
If Q₁, Q₂, Q₃, Q₄ are the quadrants in a Cartesian plane then Q₂ ∩ Q₃ is ……..
(a) Q₁ U Q₂
(b) Q₂ U Q₃
(C) Null set
(d) Negative x-axis
Solution:
(c) Null set

Question 10.
The distance between the point (5, -1 ) and the origin is ………
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(c) \(\sqrt{26}\)
Hint:

Question 11.
The coordinates of the point C dividing the line segment joining the points P (2, 4) and Q (5, 7) internally in the ratio 2 : 1 is ……..
(a) (\(\frac{7}{2}, \frac{11}{2}\))
(b) (3, 5)
(c) (4, 4)
(d) (4, 6)
Solution:
(d) (4, 6)
Hint:
A line divides internally in the ratio m : n
m = 2, n = 1
x₁ = 2, x₂ = 5
y₁ = 4, y₂ = 7

= (4, 6)

Question 12.
If p (\(\frac{a}{3}, \frac{b}{2}\)) is the mid-point of the line segment joining A (-4, 3) and B (-2, 4) then (a, b) is ………
(a) (-9, 7)
(b) (-3, \(\frac{7}{2}\))
(c) (9, -7)
(d) (3, –\(\frac{7}{2}\))
Solution:
(a) (-9, 7)
Hint:
Mid point of a line

\(\frac{a}{3}\) = -3 ⇒ a = -9
\(\frac{b}{2}\) = \(\frac{7}{2}\) ⇒ b = 7
(a, b) is (-9, 7)

Question 13.
In what ratio does the point Q (1, 6) divide the line segment joining the points P (2, 7) and R(-2, 3) ………
(a) 1 : 2
(6) 2 : 1
(c) 1 : 3
(d) 3 : 1
Solution:
(c) 1 : 3
Hint:
A line divides internally in the ratio m : n the point P =
(\(\frac{mx_{2}+nx_{1}}{m+n}\), \(\frac{my_{2}+ny_{1}}{m+n}\))

(1, 6) = (\(\frac{-2m+2n}{m+n}\), \(\frac{3m+7n}{m+n}\))
\(\frac{-2m+2n}{m+n}\) = 1
-2m + 2n = m + nx
-3m = n – 2n
3m = n
\(\frac{m}{n}\) = \(\frac{1}{3}\)
∴ m : n = 1 : 3
and
\(\frac{3m+7n}{m+n}\) = 6
3m + 7n = 6m + 6n
6m – 3n = 7n – 6n
3m = n
\(\frac{m}{n}\) = \(\frac{1}{3}\)

Question 14.
If the coordinates of one end of a diameter of a circle is (3, 4) and the coordinates of its centre is (-3, 2), then the coordinate of the other end of the diameter is ……..
(a) (0, -3)
(b) (0, 9)
(c) (3, 0)
(d) (-9, 0)
Solution:
(d) (-9, 0)
Hint:
Let the other end of the diameter be (a, b)

Mid point of a line =
(\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
(-3, 2) = \(\frac{3+a}{2}, \frac{4+b}{2}\)
\(\frac{3+a}{2}\) = -3
3 + a = -6
a = -6 – 3 = -9
and
\(\frac{4+b}{2}\) = 2
4 + b = 4
b = 4 – 4 = 0
The other end is (-9, 0)

Question 15.
The ratio in which the x-axis divides the line segment joining the points A (a₁, b₁) and B (a₂, b₂) is ……..
(a) b₁ : b₂
(b) -b₁ : b₂
(c) a₁ : a₂
(d) -a₁ : a₂
Solution:
(b) -b₁ : b₂
Hint:
A line divides internally in the ratio m : n the point P is,
(\(\frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}\))

The point P is (a, 0) = (\(\frac{ma_{2}+na_{1}}{m+n}, \frac{mb_{2}+nb_{1}}{m+n}\))
∴ \(\frac{mb_{2}+nb_{1}}{m+n}\) = 0
mb₂ + nb₁ = 0 ⇒ mb₂ = -nb₁
\(\frac{m}{n}\) = \(\frac{b_{1}}{b_{2}}\)
∴ m : n = -b₁ : b₂

Question 16.
The ratio in which the x-axis divides the line segment joining the points (6, 4) and (1, -7) is ………
(a) 2 : 3
(b) 3 : 4
(c) 4 : 7
(d) 4 : 3
Solution:
(c) 4 : 7
Hint:
A line divides internally in the ratio m : n the point P

-7m + 4n = 0
4n = 7m
\(\frac{m}{n}\) = \(\frac{4}{7}\)
The ratio is 4 : 7

Question 17.
If the coordinates of the mid-points of the sides AB, BC and CA of a triangle are (3, 4), (1, 1) and (2, -3) respectively, then the vertices A and B of the triangle are ……….
(a) (3, 2), (2, 4)
(b) (4, 0), (2, 8)
(c) (3, 4), (2, 0)
(d) (4, 3), (2, 4)
Solution:
(b) (4, 0), (2, 8)
Hint:

x₁ + x₂ = 6 → (1)
y₁ + y₂ = 8 → (2)
x₂ + x₃ = 2 → (3)
y₂ + y₃ = 2 → (4)
x₁ + x₃ = 4 → (5)
y₁ + y₃ = -6 → (6)
Adding (1) + (3) + (5) we get,
2(x₁ + x₂ + x₃) = 3
x₁ + x₂ + x₃ = 6
x₁ + x₃ = 4
∴ x₂ = 6 – 4 = 2
x₂ + x₃ = 2
x₁ = 6 – 2 = 4
Adding (2) + (4) + (6) we get,
2 (y₁ + y₂ + y₃) = 4
y₁ + y₂ + y₃ = 2
y₂ + y₃ = 2
∴ y₁ = 2 – 2 = 0
y₁ + y₃ = -6
y₂ = 2 + 6 = 8
∴ The vertices A is (4, 0) and B is (2, 8).

Question 18.
The mid-point of the line joining (-a, 2b) and (-3a, -4b) is ……..
(a) (2a, 3b)
(b) (-2a, -b)
(c) (2a, b)
(d) (-2a, -3b)
Solution:
(b) (-2a, -b)
Mid points of line

= (-2a, -b)

Question 19.
In what ratio does the y-axis divides the line joining the points (-5, 1) and (2, 3) internally ………
(a) 1 : 3
(b) 2 : 5
(c) 3 : 1
(d) 5 : 2
Solution:
(d) 5 : 2
Hint:
When it cut the y-axis the point P is (0, a)
A line divides internally in the ratio m : n the point

2m – 5n = 0 ⇒ 2m = 5n
\(\frac{m}{n}\) = \(\frac{5}{2}\) ⇒ m : n = 5 : 2

Question 20.
If (1, -2), (3, 6), (x, 10) and (3, 2) are the vertices of the parallelogram taken in order, then the value of x is ………
(a) 6
(b) 5
(c) 4
(d) 3
Solution:
(b) 5
Hint:
Since ABCD is a parallelogram

Mid-point of AC = Mid-point of BD

\(\frac{1+x}{2}\) = \(\frac{6}{2}\) ⇒ 1 + x = 6 ⇒ x = 6 – 1 = 5
The value of x = 5

Students can download Maths Chapter 6 Trigonometry Ex 6.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.4

Question 1.
Find the value of the following:
(i) sin 49°
(ii) cos 74° 39′
(iii) tan 54° 26′
(iv) sin 21° 21′
(v) cos 33° 53′
(vi) tan 70° 17′
Solution:
(i) sin 49° = 0.7547

(ii) cos 74° 39′ = cos 74° 39′ + 3′
From the table we get,
cos 74° 36′ = 0.2656
Mean difference of 3 = 8
cos 74° 39′ = 0.2656 – 8
= 0.2648

(iii) tan 54° 26′ = tan 54° 24′
tan 54° 24′ = 1.3968
Mean difference of 2 = 17
tan 54° 26’ = 1.3968 + 17
= 1.3985

(iv) sin 21° 21′ = sin 21° 18’+ 3’
sin 21° 18’ = 0.3633
Mean difference of 3 = 8(+)
sin 21° 21′ = 0.3633 + 8
= 0.3641

(v) cos 33° 53′ = cos 33° 48’ + 5′
cos 33° 48′ = 0.8310
Mean difference of 5 = 8
cos 33° 53′ = 0.8310 – 8
= 0.8302

(vi) tan 70° 17′ = tan 70° 12’+ 5’
tan 70° 12′ = 2. 7776
Mean difference of 5 = 131
tan 70° 17′ = 2. 7776 + 131
= 2.7907

Question 2.
Find the value of θ if
(i) sin θ = 0.9975
(ii) cos θ = 0.6763
(iii) tan θ = 0.0720
(iv) cos θ = 0.0410
(v) tan θ = 7.5958
Solution:
(i) sin θ = 0.9975
From the table we get,
= 0.9974 + 0.0001
= 85° 54′ + 1′
= 85° 55′ (or) 85° 56′ (or) 85° 57

(ii) cos θ = 0.6763
= 0. 6769 – 0.0006′
= 47° 24′ + 3′
= 47° 27

(iii) tan θ = 0. 0720
= 0. 0717 + 0.0003
= 4° 6′ + 1′
= 4° 7′

(iv) cos θ = 0.0410
= 0.0419 – 0.0009
= 87° 36′ + 3′
= 87° 39′

(v) tan θ = 7. 5958
= 7. 5958 + 0 (from the natural table)
= 82° 30′ (tangent table)

Question 3.
Find the value of the following:
(i) sin 65° 39′ + cos 24° 57’ + tan 10° 10′
(ii) tan 70° 58′ + cos 15° 26′ – sin 84° 59′
Solution:
(i) sin 65° 39′ = 0.9111
cos 24° 57′ = 0.9066
tan 10° 10′ = 0.1793
sin 65° 39′ + cos 24° 57′ + tan 10° 10′
= 0.9111 + 0.9066 + 0.1793
= 1.9970
sin 65° 39′ + cos 24° 57′ + tan 10° 10′ = 1.9970

(ii) tan 70° 58′ = 2. 8982
cos 15° 26′ = 0. 9639
sin 84° 59′ = 0.9962
tan 70° 58′ + cos 15° 26′ – sin 84° 59′
= 2. 8982 + 0. 9639 – 0.9962
= 3.8621 – 0.9962
= 2.8659

Question 4.
Find the area of a right triangle whose hypotenuse is 10 cm and one of the acute angle is 24°24′.
Solution:
In the ΔABC,

sin 24° 24′ = \(\frac{AB}{AC}\)
0. 4131 = \(\frac{AB}{10}\)
∴ AB = 4.131 cm
In the ΔABC,
cos 24° 24′ = \(\frac{BC}{AC}\)
0.9107 = \(\frac{BC}{10}\)
∴ BC = 9.107 cm
Area of the right angle = \(\frac{1}{2}\) × BC × AB
= \(\frac{1}{2}\) × 9.107 × 4.131
= \(\frac{37.62}{2}\)
Area of the triangle = 18.81 cm²

Question 5.
Find the angle made by a ladder of length 5m with the ground, if one of its end is 4m away from the wall and the other end is on the wall.
Solution:
AC is the length of the ladder.

In the right ΔABC,
cos θ = \(\frac{BC}{AC}\)
cos θ = \(\frac{4}{5}\)
= 0.8
cos θ = 0.8000
θ = 36° 52′
Angle made by a ladder with the ground is 36° 52′

Question 6.
In the given figure, HT shows the height of a tree standing vertically. From a point P, the angle of elevation of the top of the tree (that is ∠P) measures 42° and the distance to the tree is 60 metres. Find the height of the tree.
Solution:
Let the height of the tree HT be “x”

In the ΔHTP,
tan 42° = \(\frac{HT}{PT}\)
0.9004 = \(\frac{x}{60}\)
x = 0.9004 × 60
= 54.024
The height of the tree = 54.02 m

Students can download Maths Chapter 3 Algebra Ex 3.12 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.12

Question 1.
Solve by the method of elimination
(i) 2x – y = 3; 3x + y = 7
Solution:
2x – y = 3 → (1)
3x + y = 7 → (2)
By adding (1) and (2)
5x + 0 = 10
x = \(\frac{10}{5}\)
x = 2
Substitute the value of x = 2 in (1)
2(2) – y = 3
4 – y = 3
-y = 3 – 4
-y = -1
y = 1
The value of x = 2 and y = 1

(ii) x – y = 5; 3x + 2y = 25
Solution:
x – y = 5 → (1)
3x + 2y = 25 → (2)
(1) × 2 ⇒ 2x – 2y = 10 → (3)
(2) × 1 ⇒ 3x + 2y = 25 → (2)
(3) + (2) ⇒ 5x + 0 = 35
x = \(\frac{35}{5}\)
= 7
Substitute the value of x = 7 in (1)
x – y = 5
7 – y = 5
-y = 5 – 7
-y = -2
y = 2
∴ The value of x = 7 and y = 2

(iii) \(\frac{x}{10}\) + \(\frac{y}{5}\) = 14; \(\frac{x}{8}\) + \(\frac{y}{6}\) = 15
Solution:
\(\frac{x}{10}\) + \(\frac{y}{5}\) = 14
LCM of 10 and 5 is 10
Multiply by 10
x + 2y = 140 → (1)
\(\frac{x}{8}\) + \(\frac{y}{6}\) = 15
LCM of 8 and 6 is 24
3x + 4y = 360 → (2)
(1) × 2 ⇒ 2x + 4y = 280 → (3)
(2) × 1 ⇒ 3x + 4y = 360 → (2)
(3) – (2) ⇒ -x + 0 = -80
∴ x = 80
Substitute the value of x = 80 in (1)
x + 2y = 140
80 + 2y = 140
2y = 140 – 80
2y = 60
y = \(\frac{60}{2}\)
y = 30
∴ The value of x = 80 and y = 30

(iv) 3(2x + y) = 7xy; 3(x + 3y) = 11xy
Solution:
3(2x + y) = 7xy
6x + 3y = 7xy
Divided by xy

3a + 6b = 7 → (1)
3(x + 3y) = 11xy
3x + 9y = 11xy
Divided by xy

9a + 3b = 11 → (2)
(1) × 3 ⇒ 9a + 18b = 21 → (3)
(2) × 1 ⇒ 9a + 3b = -11 → (2)
(3) – (2) ⇒ 15b = 10
b = \(\frac{10}{15}\) = \(\frac{2}{3}\)
Substitute the value of b = \(\frac{2}{3}\) in (1)
3a + 6 × \(\frac{2}{3}\) = 7
3a + 4 = 7
3a = 7 – 4
3a = 3
a = \(\frac{3}{3}\)
= 1
But \(\frac{1}{x}\) = a
\(\frac{1}{x}\) = 1
x = 1
But \(\frac{1}{y}\) = b
\(\frac{1}{y}\) = \(\frac{2}{3}\)
2y = 3
y = \(\frac{3}{2}\)
∴ The value of x = 1 and y = \(\frac{3}{2}\)

(v) \(\frac{4}{x}\) + 5y = 7; \(\frac{3}{x}\) + 4y = 5
Solution:
Let \(\frac{1}{x}\) = a
4a + 5y = 7 → (1)
3a + 4y = 5 → (2)
(1) × 4 ⇒ 16a + 20y = 28 →(3)
(2) × 5 ⇒ 15a + 20y = 25 → (4)
(3) – (4) ⇒ a + 0 = 3
a = 3
Substitute the value of a = 3 in (1)
4(3) + 5y = 7
5y = 7 – 12
5y = -5
5y = \(\frac{-5}{5}\) = -1
But \(\frac{1}{x}\) = a
\(\frac{1}{x}\) = 3
3x = 1 ⇒ x = \(\frac{1}{3}\)
The value of x = \(\frac{1}{3}\) and y = -1

(vi) 13x + 11y = 70; 11x + 13y = 74
Solution:
13x + 11y = 70 → (1)
11x + 13y = 74 → (2)
(1) + (2) ⇒ 24x + 24y = 144
x + y = 6 → (3) (Divided by 24)
(1) – (2) ⇒ 2x – 2y = -4
x – y = -2 → (4) (Divided by 2)
(4) + (3) ⇒ 2x = 4
x = \(\frac{4}{2}\)
= 2
Substitute the value x = 2 in (3)
2 + y = 6
y = 6 – 2
= 4
∴ The value of x = 2 and y = 4

Question 2.
The monthly income of A and B are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 5,000 per month, find the monthly income of each.
Solution:
Let the income of “A” be “x” and the income of “B” be “y”.
By the given first condition
x : y = 3 : 4
4x = 3y (Product of the extreme is equal to the product of the means)
4x – 3y = 0 → (1)
Expenditure of A = x – 5000
Expenditure of B = y – 5000
By the given second condition
x – 5000 : y – 5000 = 5 : 7
7(x – 5000) = 5(y – 5000)
7x – 35000 = 5y – 25000
7x – 5y = -25000 + 35000
7x – 5y = 10000 → (2)
(1) × 5 ⇒ 20x – 15y = 0 → (3)
(2) × 3 ⇒ 21x – 15y = 30000 → (4)
(3) – (4) ⇒ x + 0 = 30000
x = 30000
Substitute the value of x in (1)
4 (30000) – 3y = 0
120000 = 3y
y = \(\frac{120000}{3}\) = 40000
∴ Monthly income of A is Rs 30,000
Monthly income of B is Rs 40,000

Question 3.
Five years ago, a man was seven times as old as his son, while five year hence, the man will be four times as old as his son. Find their present age.
Solution:
Let the age of a man be “x” and the age of a son be “y”
5 years ago
Age of a man = x – 5 years
Age of his son = y – 5 years
By the given first condition
x – 5 = 7(y – 5)
x – 5 = 7y – 35
x – 7y = -35 + 5
x – 7y = -30 → (1)
Five years hence
Age of a man = x + 5 years
Age of his son = y + 5 years
By the given second condition
x + 5 = 4 (y + 5)
x + 5 = 4y + 20
x – 4y = 20 – 5
x – 4y = 15 → (2)
(1) – (2) ⇒ -3y = -45
3y = 45
y = \(\frac{45}{3}\)
= 15
Substitute the value of y = 15 in (1)
x – 7(15) = -30
x – 105 = -30
x = -30 + 105
= 75
Age of the man is 75 years
Age of his son is 15 years