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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.7

Question 1.
Solve the following system of homogeneous equations.
(i) 3x + 2y + 7z = 0, 4x – 3y – 2z = 0, 5x + 9y + 23z = 0
(ii) 2x + 3y – z = 0, x – y – 2z = 0, 3x + y + 3z = 0
Solution:
(i) The matrix form of the above equations is

The above matrix is in echelon form.
Here ρ(A, B) = ρ(A) < number of unknowns.
The system is consistent with infinite number of solutions. To find the solutions.
Writing the equivalent equations.
We get 3x + 2y + 7z = 0 ……. (1)
-17y – 34z = 0 ……. (2)
Taking z = t in (2) we get -17y – 34t = 0
⇒ -17y = 34t
⇒ y = -2t
Taking z = t, y = -2t in (1) we get
3x + 2(-2t) + 7t = 0
⇒ 3x – 4t + 7t = 0
⇒ 3x = -3t
⇒ x = -t
So the solution is x = -t; y = -2t; and z = t, t ∈ R
(ii) The matrix form of the equations is
\(\left(\begin{array}{ccc}{2} & {3} & {-1} \\ {1} & {-1} & {-2} \\ {3} & {1} & {3}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right)\)
(i.e) AX = B
The augmented matrix [A, B] is

The above matrix is in echelon form also ρ(A, B) = ρ(A) = 3 = number of unknowns
The system is consistent with unique solution, x = y = z = 0
(i.e) The system has trivial solution only.

Question 2.
Determine the values of λ for which the following system of equations.
x + y + 3z = 0, 4x + 3y + λz = 0, 2x +y + 2z = 0 has
(i) a unique solution
(ii) a non-trivial solution
Solution:
The matrix form of the equation is
\(\left(\begin{array}{lll}{1} & {1} & {3} \\ {4} & {3} & {\lambda} \\ {2} & {1} & {2}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right)\)
(i.e) AX = B
The augmented matrix [A, B] is

The above matrix is in echelon form
Case 1: When λ ≠ 8, ρ(A, B) = ρ(A) = 3 = number of unknowns
The system is consistent with a unique solution.
Case 2: When λ = 8, ρ(A, B) = ρ(A) = 2 < number of unknowns.
The system is consistent with non-trivial solutions.

Question 3.
By using the Gaussian elimination method, balance the chemical reaction equation:
C₂H₆ + O₂ → H₂O + CO₂
Solution:
We are searching for positive integers x₁, x₂, x₃ and x₄ such that
x₁C₂H₆ + x₂O₂ = x₃H₂O + x₄CO₂ ……. (1)
The number of carbon atoms on LHS of (1) should be equal to the number of carbon atoms on the RHS of (1).
So we get a linear homogeneous equation.
2x₁ = x₄
⇒ 2x₁ – x₄ = 0 …… (2)
Similarly considering hydrogen and oxygen atoms we get respectively.
2x₂ = x₃ + 2x₄
⇒ 2x₂ – x₃ – 2x₄ = 0 …… (3)
and -2x₃+ 3x₄ = 0 …… (4)
Equations (2), (3) and (4) constitute a homogeneous system of linear equations in four unknowns.
The augmented matrix (A, B) is

Now ρ(A, B) = ρ(A) = 3 < number of unknowns.
So the system is consistent and has an infinite number of solutions.
Writing the equations using the echelon form we get
2x₁ – x₄ = 0 …… (5)
2x₂ – x₃ – 2x₄ = 0 ……. (6)
-2x₃ + 3x₄ = 0 ……. (7)

Since x₁, x₂, x₃ and x₄ are positive integers. Let us choose t = 4t.
Then we get x₁ = 2, x₂ = 7, x₃ = 6, and x₄ = 4
So the balanced equation is 2C₂H₆ + 7O₂ → 6H₂O + 4CO₂.

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.7 Additional Problems

Question 1.
Solve the following homogeneous linear equations.
x + 2y – 5z = 0,
3x + 4y + 6z = 0,
x + y + z = 0
Solution:
The given system of equations can be written in the form of matrix equation

AX = B
The augmented matrix is

This is in the echelon form.
Clearly ρ[A, B] = 3 and ρ(A) = 3
∴ ρ(A) = ρ[A, B] = 3 = number of unknowns
∴ The given system of equations is consistent and has a unique solution. i.e., trivial solution
∴ x = 0, y = 0 and z = 0
Note: Since ρ(A) = 3, | A | ≠ 0 i.e. A is non-singular;
∴ The given system has only trivial solution x = 0, y = 0, z = 0

Question 2.
For what value of n the equations.
x + y + 3z = 0,
4x + 3y + µz = 0,
2x + y + 2z = 0 have a
(i) trivial solution,
(ii) non-trivial solution.
Solution:
The system of equations can be written as AX = B

Case (i): If µ ≠ 8 then 8 – µ ≠ 0 and hence there are three non-zero rows.
∴ ρ[A] = ρ[A, B] = 3 = the number of unknowns.
∴ The system has the trivial solution x = 0, y = 0, z = 0
Case (ii): If µ = 8 then.
ρ[A, B] = 2 and ρ(A) = 2
∴ ρ(A) = ρ[A, B] = 2 < number of unknowns.
The given system is equivalent to
x + y + 3z = 0; y + 4z = 0
∴ y= – 4z ; x = z
Taking z = k, we get x = k,y = – 4k, z = k [k ∈ R – {0}] which are non-trivial solutions.
Thus the system is consistent and has infinitely many non-trivial solutions.
Note: In case (ii) the system also has trivial solution. For only non-trivial solutions we removed k = 0.

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.11

Question 1.

Solution:

Question 2.
A man standing directly opposite to one side of a road of width x meter views a circular shaped traffic green signal of diameter a meter on the other side of the road. The bottom of the green signal is b meter height from the horizontal level of viewer’s eye. If a denotes the angle subtended by the diameter of the green signal at the viewer’s eye, then prove that

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.11 Additional Questions

Question 1.

Solution:

Question 2.

Solution:

The negative value of x is rejected since it makes RHS negative
∴ x = \(\frac{1}{6}\)

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.6

Question 1.
Test for consistency and if possible, solve the following systems of equations by rank method.
(i) x – y + 2z = 2, 2x + y + 4z = 7, 4x – y + z = 4
(ii) 3x + y + z = 2, x – 3y + 2z = 1, 7x – y + 4z = 5
(iii) 2x + 2y + z = 5, x – y + z = 1, 3x + y + 2z = 4
(iv) 2x – y + z = 2, 6x – 3y + 3z = 6, 4x – 2y + 2z = 4
Solution:
(i) Here the number of unknowns = 3.
The matrix form of the system is AX = B where

(i.e) AX = B
The augmented matrix (A, B) is
\((\mathrm{A}, \mathrm{B})=\left(\begin{array}{cccc}{1} & {-1} & {2} & {2} \\ {2} & {1} & {4} & {7} \\ {4} & {-1} & {1} & {4}\end{array}\right)\)
Applying Gaussian elimination method on [A,B] we get

The above matrix is in echelon form also ρ(A, B) = ρ(A) = 3 = number of unknowns
The system of equations is consistent with a unique solution. To find the solution.
Now writing the equivalent equations we get
\(\left(\begin{array}{ccc}{1} & {-1} & {2} \\ {0} & {3} & {0} \\ {0} & {0} & {-7}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{r}{2} \\ {3} \\ {-7}\end{array}\right)\)
x – y + 2z = 2
3y = 3 ⇒ y = 1
7z = -7 ⇒ z = 1
Substituting z = y = 1 in (1) we get
x – 1 + 2 = 2 ⇒ x = 1
⇒ x = y = z = 1
(ii) Here the number of unknowns is 3.
The matrix form of the given system of equations is:
\(\left(\begin{array}{ccc}{3} & {1} & {1} \\ {1} & {-3} & {2} \\ {7} & {-1} & {4}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{2} \\ {1} \\ {5}\end{array}\right)\)
AX = B
(i.e) Now the augmented matrix [A, B] is

The above matrix is in echelon form also
ρ(A, B) = ρ(A) = 2< number of unknowns
The system of equations is consistent with the infinite number of solutions.
To find the solution:
Now writing the equivalent equations we get


(iii) Here the number of unknowns is 3.
The matrix form of the given equation is
\(\left(\begin{array}{ccc}{2} & {2} & {1} \\ {1} & {-1} & {1} \\ {3} & {1} & {2}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{5} \\ {1} \\ {4}\end{array}\right)\)
AX = B

The above matrix is in echelon form.
Here ρ(A, B) = 3; ρ(A) = 2
So ρ(A, B) ≠ ρ(A)
The system of equations is inconsistent with no solution.
(iv) Here the number of unknowns is 3.
The matrix form of the given equation is
\(\left(\begin{array}{ccc}{2} & {-1} & {1} \\ {6} & {-3} & {3} \\ {4} & {-2} & {2}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{2} \\ {6} \\ {4}\end{array}\right)\)
AX = B
The augmented matrix [A, B] is

The above matrix is in echelon form also ρ(A, B) = ρ(A) = 1 < number of unknowns
The system of equations is consistent with the infinite number of solutions.
To find the Solution
Now writing the equivalent equations we get

Question 2.
Find the value of k for which the equations kx – 2y + z = 1, x – 2ky + z = -2, x – 2y + kz = 1 have
(i) no solution
(ii) unique solution
(iii) infinitely many solution
Solution:
The matrix form of the given system of equation is
\(\left(\begin{array}{ccc}{k} & {-2} & {1} \\ {1} & {-2 k} & {1} \\ {1} & {-2} & {k}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{c}{1} \\ {-2} \\ {1}\end{array}\right)\)
(i.e) AX = B
The augmented matrix (A, B) is


k² + k – 2 = (k + 2) (k – 1)
The above matrix is in echelon form
Case 1: when k = 1, ρ(A, B) = 3, ρ(A) = 2 (i.e) ρ (A, B) ≠ ρ (A)
The system is inconsistent and the system has no solution.
Case 2: when k ≠ 1, k ≠ -2, then ρ(A, B) = ρ(A) = 3 = number of unknowns
The system is consistent with unique solution.
Case 3: When k = -2 then ρ(A) = ρ(A, B) = 2 < number of unknowns.
The system is consistent with the infinite number of solutions.

Question 3.
Investigate the values of λ and µ the system of linear equations.
2x + 3y + 5z = 9, 7x + 3y – 5z = 8, 2x + 3y + λz = µ, have
(i) no solution
(ii) a unique solution
(iii) an infinite number of solutions.
Solution:
The matrix form of the above equation is
\(\left(\begin{array}{ccc}{2} & {3} & {5} \\ {7} & {3} & {-5} \\ {2} & {3} & {\lambda}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{9} \\ {8} \\ {\mu}\end{array}\right)\)
(i.e) AX = B
The augmented matrix [A, B] is

The above matrix is in echelon form
Case 1: When λ = 5, μ ≠ 9
ρ(A) = 2, ρ(A, B) = 3 (i.e) ρ(A, B) ≠ p(A)
The system is inconsistent and it has no solution.
Case 2: When λ ≠ 5, μ ∈ R,
ρ(A, B) = ρ(A) = 3 = number of unknowns
The system is consistent with a unique solution.
Case 3: When λ = 5, μ = 9,
then ρ(A, B) = ρ(A) = 2 < number of unknowns
The system is consistent with infinite number of solutions.

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.6 Additional Problems

Question 1.
Discuss the solutions of the system of equations for all values of λ.
x + y + z = 2,
2x + y – 2z = 2,
λx + y + 4z = 2
Solution:
The augmented matrix is

Case 1:
If λ ≠ 0, then we get ρ(A, B) = ρ(A) = 3 = number of unknowns.
∴ The system has unique solution.
Case 2: If λ = 0,then ρ(A, B) = ρ(A) = 2 < the number of unknowns ⇒ the system has more number of solutions. Taking λ = 0 we get the systems of equations as
x + 4z = 2 ……..(1)
2y – 6z = 0 …….. (2)
Taking z = k in (2), we get
2y – 6k = 0 ⇒ 2y = 6k
y = 3k
Taking z = A in (1) we get,
x + 4k = 2 ⇒ x = 2 – 4k
∴ The solution is x = 2 – 4 k, y = 3 k, z = k

Question 2.
For what values of k, the system of equations kx + y + z = 1, x + ky + z = 1, x + y + kz = 1 have
(i) unique solution,
(ii) more than one solution,
(iii) no solution.
Solution:
The augmented matrix is

Case (i) : When k ≠ 1, the system has a unique solution
Case (ii) : When k = 1, the system reduces to a single equation x + y + z = 1. The system can have more than one solution.
Case (iii) : When k = -2, the system is inconsistent
∴ ρ(A, B) ≠ ρ(A); ∴ the system has no solution.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.2

Question 1.
Evaluate the following if z = 5 – 2i and w = -1 + 3i
(i) z + w
(ii) z – iw
(iii) 2z + 3w
(iv) zw
(v) z² + 2zw+ w²
(vi) (z + w)²
Solution:
(i) z = 5 – 2i, w = -1 + 3i
z + w = (5 – 2i) + (-1 + 3i)
= (5 – 1) + (-2i + 3i)
= 4 + i
(ii) z – iw = (5 – 2i) – i (-1 + 3i)
= 5 – 2i + i + 3
= (5 + 3) + (-2i + i)
= 8 – i
(iii) 2z + 3w = 2(5 – 2i) + 3 (-1 +3i)
= 10 – 4i – 3 + 9i
= 7 + 5 i
(iv) zw = (5 – 2i) (-1 + 3i)
= -5 + 15i + 2i – 6i²
= -5 + 17i + 6
= 1 + 17i
(v) z² + 2zw + w² = (z + w)² [from (i)]
= (4 + i)²
= 16 – 1 + 8i
= 15 + 8i
(vi) (z + w)² = 15 + 8z [from (v)]

Question 2.
Given the complex number z = 2 + 3i, represent the complex numbers in Argand diagram.
(i) z, iz, and z + iz
(ii) z, -iz, and z – iz
Solution:
(i) z = 2 + 3i
iz = i(2 + 3i)
= (2i – 3)
= -3 + 2i
z + iz = (2 + 3i) + (-3 + 2i)
= -1 + 5i

(ii) z = 2 + 3i
-iz = -i(2 + 3i)
= -2i – 3i²
= (3 – 2i)
z – iz = (2 + 3i) + (3 – 2i)
= 5 + i

Question 3.
Find the values of the real numbers x and y, if the complex numbers.
(3 – i) x – (2 – i) y + 2i + 5 and 2x + (-1 + 2i) y + 3 + 2i are equal
Solution:
(3 – i) x – (2 – i) y + 2i + 5 = 2x + (-1 + 2i) y + 3 + 2i
⇒ 3x – ix – 2y + iy + 2i + 5 = 2x – y + 2yi + 3 + 2i
⇒ (3x – 2y + 5) + 1 (-x + y + 2) = (2x – y + 3) + i (2y + 2)
Equate real parts on both sides
3x – 2y + 5 = 2x – y + 3
x – y = -2 ……. (1)
Equate imaginary parts on both sides
-x + y + 2 = 2y + 2
-x – y = 0
x + y = 0 ……. (2)
(1) + (2) ⇒ 2x = -2
x = -1
Substituting x = -1 in (2)
-1 + y = 0
⇒ y = 1
∴ x = -1, y = 1

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.2 Additional Problems

Question 1.
Find the real values of x and y, if
(i) (3x – 7) + 2 iy = -5y + (5 + x)i
(ii) (1 – i)x + (1 + i)y = 1 – 3i
(iii) (x + iy)(2 – 3i) = 4 + i
(iv)
Solution:
(i) We have (3x – 7) + 2 iy = 5y + (5 + x)i
⇒ 3x – 7 = 5y and 2y = 5 + x
⇒ 3x + 5y = 7 and x – 2y = -5
⇒ x = -1 y = 2

(ii) We have, (1 – i) x + (1 + i)y = 1 – 3i
⇒ (x + y) + i(-x + y) = 1 – 3i
⇒ x + y = 1
and -x + y = 3 [On equating real and imaginary parts]
⇒ x = 2 and y = -1

⇒ x + y – 2 = 0 and y – x = 10
⇒ x = -4 , y = 6.

Question 2.
Find the real values of x and y for which the complex numbers -3 + ix²y and x² + y + 4i are conjugate of each other.
Solution:
Since -3 + ix²y and x² + y + 4i are complex conjugates.
∴ -3 + ix²y = x² + y + 4i
… (0 and, x² y = -4 …. (ii)

Question 3.
Given x = 2 – 3i and y = 4 + i

Solution:
Do it yourself

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.1

Simplify the following:

Question 1.
i¹⁹⁴⁷ + i¹⁹⁵⁰
Solution:

Question 2.
i¹⁹⁴⁸ – i^-1869
Solution:

Question3.
\(\sum_{n=1}^{12} i^{n}\)
Solution:

Question 4.
\(i^{59}+\frac{1}{i^{59}}\)
Solution:

Question 5.
i i² i³ …….. i²⁰⁰⁰
Solution:

Question 6.
\(\sum_{n=1}^{10} i^{n+50}\)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 2.1 Additional Problems

Question 1.
Evaluate the following:

Solution:
135 leaves remainder as 3 when it is divided by 4.
∴ i¹³⁵ = i³ = -i

(ii) The remainder is 3 when 19 is divided by 4.
∴ i¹⁹ = i³ = -i

(iii) We have, i⁹⁹⁹ = 1/i⁹⁹⁹
On dividing 999 by 4, we obtain 3 as the remainder.

(iv)

Question 2.
Show that:

Solution:

Question 3.

Solution:

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.10

Question 1.
Determine whether the following measurements produce one triangle, two triangles or no triangle:
∠B = 88°, a = 23, b = 2. Solve if solution exists.
Solution:
We are given a = 23,
b = 2, and
∠B = 88°.
So we can

Question 2.
If the sides of a ∆ABC are a = 4, b = 6 and c = 8, then show that 4 cos B + 3 cos C = 2.
Solution:
a = 4,
b = 6,
c = 8
To prove 4 cos B + 3 cos C = 2

Question 3.
In a ∆ABC, if a = \(\sqrt{3}\) – 1, b = \(\sqrt{3}\) + 1 and C = 60°, find the other side and other two angles.
Solution:

Question 4.

Solution:

Question 5.
In a ∆ABC, if a = 12 cm, b = 8 cm and C = 30°, then show that its area is 24 sq.cm.
Solution:
a = 12 cm,
b = 8 cm,
C = 30°

Question 6.
In a ∆ABC, if a = 18 cm, b = 24 cm and c = 30 cm, then show that its area is 216 sq.cm.
Solution:
a = 18 cm,
b = 24 cm,
c = 30 cm
The sides form a right angled triangle

Question 7.
Two soldiers A and B in two different underground bunkers on a straight road, spot an intruder at the top of a hill. The angle of elevation of the intruder from A and B to the ground level in the eastern direction are 30° and 45° respectively. If A and B stand 5 km apart, find the distance of the intruder from B.
Solution:
By using sine formula
\(\frac{x}{\sin 30^{\circ}}=\frac{5}{\sin 15^{\circ}}\)

Question 8.
A researcher wants to determine the width of a pond from east to west, which cannot be done by actual measurement. From a point P, he finds the distance to the eastern-most point of the pond to be 8 km, while the distance to the western most point from P to be 6 km. If the angle between the two lines of sight is 60°, find the width of the pond.
Solution:
p² = W² + E² – 2WE cos P
P² = 64 + 36 – 2 × 8 × 6 × Cos 60°

Question 9.
Two Navy helicopters A and B are flying over the Bay of Bengal at same altitude from the sea level to search a missing boat. Pilots of both the helicopters sight the boat at the same time while they are apart 10 km from each other. If the distance of the boat from A is 6 km and if the line segment AB subtends 60° at the boat, find the distance of the boat from B.
Solution:

Question 10.
A straight tunnel is to be made through a mountain. A surveyor observes the two extremities A and B of the tunnel to be built from a point P in front of the mountain. If AP = 3 km, BP = 5 km and ∠APB = 120°, then find the length of the tunnel to be built.
Solution:
p² = a² + b² – 2ab cos P
p² = 9 + 25 – 30 Cos 120°
p² = 9 + 25 – 30 (-1/2) = 34 + 15 = 49

⇒ p = \(\sqrt{49}\) = 7 km

Question 11.
A farmer wants to purchase a triangular shaped land with sides 120 feet and 60 feet and the angle included between these two sides is 60°. If the land costs ? 500 per sq.ft, find the amount he needed to purchase the land. Also find the perimeter of the land.
Solution:

Question 12.
A fighter jet has to hit a small target by flying a horizontal distance. When the target is sighted, the pilot measures the angle of depression to be 30°. If after 100 km, the target has an angle of depression of 60°, how far is the target from the fighter jet at that instant?
Solution:

Question 13.
A plane is 1 km from one landmark and 2 km from another. From the planes point of view the land between them subtends an angle of 60°. How far apart are the landmarks?
Solution:

Question 14.
A man starts his morning walk at a point A reaches two points B and C and finally back to A such that ∠A= 60° and ∠B = 45°, AC = 4 km in the ∆ABC. Find the total distance he covered during his morning walk.
Solution:

Question 15.
Two vehicles leave the same place P at the same time moving along two different roads. One vehicle moves at an average speed of 60 km/hr and the other vehicle moves at an average speed of 80 km/hr. After half an hour the vehicle reach the destinations A and B. If AB subtends 60° at the initial point P, then find AB.
Solution:

= 900+ 1600 – 1200 = 1300

Question 16.
Suppose that a satellite in space, an earth station and the centre of earth all lie in the same plane. Let r be the radius of earth and R be the distance from the centre of earth to the satellite. Let d be the distance from the earth station to the satellite. Let 30° be the angle of elevation from the earth station to the satellite. If the line segment connecting earth station and satellite subtends angle α at the centre of earth, then prove that

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.10 Additional Questions Solved

Question 1.
Given a = 8, b = 9, c = 10, find all the angles.
Solution:

Question 2.
Given a = 31, b = 42, c = 57, find all the angles.
Solution:
Since the sides are larger quantities, use half angles formulae

Question 3.
In a triangle ABC, A = 35° 17′ ; C = 45° 13′ ; b = 42.1 Solve the triangle
Solution:
The unknown parts are B, a, c,
B = 180 – (A + C) = 180 – (35° 17′ + 45° 13′)
= 99° 30′
To find sides, use sine formula


log c = log 42.1 + log sin 45° 31 – log sin 99° 30′
= 1.6243 + 1.8511 – 1.9940
= 1.4754 – 1.9940
= 1.4754 – [-1 + 0.9940] = 1.4814
⇒ c = 30.3°
Thus B = 99° 30′ ; a = 24.65° ; c = 30.3°

Question 4.
Solve the triangle ABC if a = 5, b = 4 and C = 68°.
Solution:
To find c, use c² = a² + b² – 2ab cos C
c² = 25 + 16 – 2 × 5 × 4 cos 68°
= 41 – 40 × 0.3746 = 26.016
c = 5.1
To find the other two angles, use sine formula.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 1.
Solve the following systems of linear equations by Gaussian elimination method:
(i) 2x – 2y + 3z = 2, x + 2y – z = 3, 3x – y + 2z = 1
(ii) 2x + 4y + 6z = 22, 3x + 8y + 5z = 27, -x + y + 2z = 2
Solution:
(i) 2x – 2y + 3z = 2, x + 2y – z = 3 and 3x – y + 2z = 1
The matrix form of the above equations is \(\left(\begin{array}{rrr}{2} & {-2} & {3} \\ {1} & {2} & {-1} \\ {3} & {-1} & {2}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{2} \\ {3} \\ {1}\end{array}\right)\)
(i.e) AX = B
The augment matrix (A, B) is

The above matrix is in echelon form.
Now writing the equivalent equations

Substituting z = 4 in (2) we get
-6y + 20 = -4
⇒ -6y = -4 – 20 = -24
⇒ y = 4
Substituting z = 4 and y = 4 in (1) we get
x + 8 – 4 = 3
⇒ x + 4 = 3
⇒ x = 3 – 4 = -1
So, x = -1; y = 4; z = 4
(ii) 2x + 4y + 6z = 22 …… (1)
3x + 8y + 5z = 27 ……. (2)
-x + y + 2z = 2 ……. (3)
Divide equation (1) by 2 we get
x + 2y + 3z = 11 ……. (1)
3x + 8y + 5z = 27 …….. (2)
-x + y + 2z = 2 ……. (3)
The matrix form of the above equations is \(\left(\begin{array}{ccc}{1} & {2} & {3} \\ {3} & {8} & {5} \\ {-1} & {1} & {2}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{c}{11} \\ {27} \\ {2}\end{array}\right)\)
(i.e) AX = B
The augment matrix (A, B) is


The above matrix is in echelon form.
Now writing the equivalent equations.

Substituting z = 2 in (2) we get
y – 4 = -3
⇒ y = -3 + 4 = 1
Substituting z = 2, y = 1 in (1) we get
x + 2(1) + 3(2) = 11
⇒ x + 2 + 6 = 11
⇒ x + 8 = 11
⇒ x = 11 – 8 = 3
x = 3, y = 1, z = 2

Question 2.
If ax² + bx + c is divided by x + 3, x – 5, and x – 1, the remainders are 21, 61 and 9 respectively. Find a, b and c. (Use Gaussian elimination method.)
Solution:
P(x) = ax² + bx + c. When P(x) is divided by x + 3, x – 5 and x – 1.
The remainders are respectively P(-3), P (5) and P (1).
We are given that P(-3) = 21; P(5) = 61; P(1) = 9
Now P(-3) = 21
⇒ a(-3)² + b(-3) + c = 21
⇒ 9a – 3b + c = 21 ……. (1)
P(5) = 61
⇒ a(5)² + b(5) + c = 61
⇒ 25a + 5b + c = 61 ……. (2)
P(1) = 9
⇒ a(1)² + b(1) + c = 9
⇒ a + b + c = 9 …… (3)
Now the matrix form of the above three equations is \(\left(\begin{array}{ccc}{9} & {-3} & {1} \\ {25} & {5} & {1} \\ {1} & {1} & {1}\end{array}\right)\left(\begin{array}{l}{a} \\ {b} \\ {c}\end{array}\right)=\left(\begin{array}{c}{21} \\ {61} \\ {9}\end{array}\right)\)
(i.e) AX = B
The augmented matrix (A, B) is

The above matrix is in echelon form now writing the equivalent equations.

Substituting c = 6 in (2) we get
-20b – 24(6) = -164
⇒ -20b = -164 + 144 = -20
⇒ b = 1
Substituting c = 6, b = 1 in (1) we get
a + 1 + 6 = 9
⇒ a = 9 – 7 = 2
So a = 2, b = 1, c = 6

Question 3.
An amount of ₹ 65,000 is invested in three bonds at the rates of 6%, 8% and 10% per annum respectively. The total annual income is ₹ 5,000. The income from the third bond is ₹ 800 more than that from the second bond. Determine the price of each bond. (Use Gaussian elimination method.)
Solution:
Let the amount invested in 6% bond be ₹ x
and the amount invested in 8% bond be ₹ y
and the amount invested in 10% bond be ₹ z
Now x + y + z = 65000 ……. (1)
\(\frac{6}{100} x+\frac{8}{100} y+\frac{10}{100} z=5000\)
(i.e) 6x + 8y + 10z = 500000
(÷ by 2) 3x + 4y + 5z = 250000 ……. (2)
Also given that \(\frac{10}{100} z-\frac{8}{100} y=800\)
(i.e) -8y + 10z = 80000
(÷ by 2) -4y + 5z = 40000 …… (3)
Now the matrix form of the above three equations is \(\left(\begin{array}{ccc}{1} & {1} & {1} \\ {3} & {4} & {5} \\ {0} & {-4} & {5}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{c}{65000} \\ {250000} \\ {40000}\end{array}\right)\)
(i.e) AX = B
Now, the augmented matrix (A, B) is

Now the above matrix is in echelon form. Writing the equivalent equations.
\(\left(\begin{array}{lll}{1} & {1} & {1} \\ {0} & {1} & {2} \\ {0} & {0} & {13}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{r}{65000} \\ {55000} \\ {260000}\end{array}\right)\)
(i.e) x + y + z = 65000
⇒ y + 2z = 55000
⇒ 13z = 260000
⇒ z = ₹ 20000
Substituting z = ₹ 20000 in (2) we get
y + 40000 = 55000
⇒ y = 55000 – 40000 = ₹ 15000
Substituting z = ₹ 20000, y = ₹ 15000 in (1) we get
x + 15000 + 20000 = 65000
⇒ x = 65000 – 35000 = ₹ 30000
So the amount invested in
6% bond x = ₹ 30000
8% bond y = ₹ 15000
and 10% bond z = ₹ 20000

Question 4.
A boy is walking along the path y = ax² + bx + c through the points (-6, 8), (-2, -12), and (3, 8). He wants to meet his friend at P(7, 60). Will he meet his friend? (Use Gaussian elimination method.)
Solution:
We are given y = ax² + bx + c
Also we are given (-6, 8), (-2, -12) and (3, 8) are points on the path.
(i) (-6, 8) is a point on y = ax² + bx + c
at x = -6, y = 8
(i.e) a(36) + b(-6) + c = 8
⇒ 36a – 6b + c = 8 …… (1)
(ii) (-2, -12) is a point on y = ax² + bx + c
at x = -2, y = -12
⇒ a(-2)² + b(-2) + c = -12
⇒ 4a – 2b + c = -12 ….. (2)
(iii) (3, 8) is a point on y = ax² + bx + c
at x = 3, y = 8
⇒ a(3)² + 6(3) + c = 8
⇒ 9a + 3b + c = 8 …… (3)
The matrix form of the above three equations is \(\left(\begin{array}{ccc}{36} & {-6} & {1} \\ {4} & {-2} & {1} \\ {9} & {3} & {1}\end{array}\right)\left(\begin{array}{l}{a} \\ {b} \\ {c}\end{array}\right)=\left(\begin{array}{c}{8} \\ {-12} \\ {8}\end{array}\right)\)
(i.e) AX = B
The augmented matrix (A, B) is

The above matrix is in echelon form. Now writing the equivalent equations we get
\(\left(\begin{array}{ccc}{36} & {-6} & {1} \\ {0} & {-12} & {8} \\ {0} & {0} & {30}\end{array}\right)\left(\begin{array}{l}{a} \\ {b} \\ {c}\end{array}\right)=\left(\begin{array}{c}{8} \\ {-116} \\ {-300}\end{array}\right)\)
(i.e) 36a – 6b + c = 8
⇒ -12b + 8c = -116
⇒ 30c = -300
⇒ c = -10
Substituting c = -10 in (2) we get
-12b + 8(-10) = -116
⇒ -12b = -116 + 80 = -36
⇒ b = 3
Substituting c = -10, b = 3 in (1) we get
36a – 6(3) + (-10) = 8
⇒ 36a – 18 – 10 = 8
⇒ 36a = 8 + 18 + 10 = 36
⇒ a = 1
a = 1, b = 3 and c = -10
y = (1)x² + (3)x + (-10)
y = x² + 3x – 10
Now at x = 7, y= (7)² + 3(7)- 10 = 49 + 21 – 10 = 60
(7, 60) is a point on the path so he will meet his friend.

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Additional Problems

Question 1.
Examine the consistency of the following system of equations. If it is consistent then solve the same.
(i) 4x + 3y + 6z = 25,
x + 5y + 7z = 13,
2x + 9y + z = 1
Solution:
The augmented matrix is

The last equivalent matrix is in the echelon form. It has three non-zero rows.

∴ The given system is consistent and has a unique solution
4x + 3y + 6z = 25 …… (1)

∴ The unique solution is x = 4, y = -1, z = 2.

Question 2.
Verify whether the given system of equations is consistent. If it is consistent, solve them. 2x + 5y + 7z = 52, x + 7 + z = 9, 2x + y – z = 0
Solution:
The given system of equations is equivalent to the single matrix equation.

AX = B
The augmented matrix is

The last equivalent matrix is in the echelon form. It has three non-zero rows.
∴ \(\rho(\mathrm{A}, \mathrm{B})=3\)

Since there are three non-zero rows, \(\rho(\mathrm{A})\) = 3

The given system is consistent and has a unique solution.
To find the solution, we see that the given system of equations is equivalent to the matrix equation.

x + y + z = 9 ……. (1)
-y – 3z = -18 ……… (2)
-4z = -20 ……… (3)
(3) ⇒ z = 5, (2) ⇒ y = 18 – 3z = 13, (1) ⇒ x = 9 – y – z ⇒ x = 9 – 3 – 5 = 1
∴ Solution is x = 1,
y = 3,
z = 5

Question 3.
Examine the consistency of the equations.
2x – 3y + 3z = 5,
3x + y – 3z = 13,
2x + 19y – 47z = 32
Solution:
The given system of equations can be written in the form of a matrix equation as

The augmented matrix is

The last equivalent matrix is in the echelon form. It has three non-zero rows.

∴ The given system is inconsistent and hence has no solution.

Question 4.
Show that the equations: x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30 are consistent and solve them.
Solution:
The matrix equation corresponding to the given system is.

The augmented matrix is

In the last equivalent matrix, there are two non-zero rows.

∴ The given system is consistent. But the value of the common rank is less than the number of unknowns. The given system has an infinite number of solutions.
The given system is equivalent to the matrix equation

x + y + z = 6 ……(1)
y + 2z = 8 ….(2)
(2) ⇒ y = 8 – 2z ; (1) ⇒ x = 6 – y – z = 6 – (8 – 2z) – z = z – 2
Taking z = k, we get x = k – 2, y = 8 – 2k; k ∈ R
Putting k = 1, we have one solution as x = – 1, y = 6, z = 1. Thus by giving different values for k we get different solutions. Hence the given system has infinite number of solutions.

You can Download Samacheer Kalvi 11th 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 11th Maths Solutions Chapter 3 Trigonometry Ex 3.9

Question 1.

Solution:

Question 2.
The angles of a triangle ABC, are in Arithmetic Progression and if b : c = \(\sqrt{3}: \sqrt{2}\), find ∠A.
Solution:

Question 3.

Solution:

⇒ a² + b² – c² = a² ⇒ b² – c² = a² – a²
⇒ b² – c² = 0 ⇒ b = c
∴ ∆ ABC is isosceles

Question 4.

Solution:

Question 5.
In a ∆ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C.
Solution:
LHS = a cos A+ 6 cos B + c cos C
Using sine formula, we get k sin A cos A + k sin B cos B + k sin C cos C k
= \(\frac{k}{2}\) [2 sin A cos A + 2 sin B cos B + 2 sin C cos C]
= \(\frac{k}{2}\) [sin 2A + sin 2B + sin 2C]
= \(\frac{k}{2}\) [2 sin (A + B) . cos (A – B) + 2 sin C . cos C]
= \(\frac{k}{2}\) [2 sin (A – B) . cos (A – B) + 2 sin C . cos C]
= \(\frac{k}{2}\) [2 sin C . cos (A – B) + 2 sin C . cos C]
= k sin C [cos(A – B) + cos C]

= k sin C [cos (A – B) – cos (A + B)]
= k sin C . 2 sin A sin B
= 2k sin A . sin B sin C
= 2a sin B sin C = RHS

Question 6.

Solution:

Question 7.
In a ∆ ABC, prove the following.

Solution:

Question 8.
In a ∆ABC, prove that (a² – b² + c²) tan B = (a² + b² – c²)tan C
Solution:

Question 9.
An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park.
Solution:
Given, the perimeter of triangular shaped park = 120 m

All sides of a triangular part would be 40 m.
i.e., a = 40 m,
b = 40 m,
c = 40 m.

Question 10.
A rope of length 12 m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed.
Solution:
The largest triangle will be an equilateral triangle

Question 11.
Derive Projection formula from
(i) Law of sines,
(ii) Law of cosines.
Solution:
(i) To Prove a = b cos c + c cos B
Using sine formula
RHS = b cos C + c cos B
= 2R sin B cos C + 2R sin C cos B
= 2R [sin B cos C + cos B sin C]
= 2R sin (B + C) = 2R [sin π – A)
= 2R sin A = a = LHS

(ii) To prove a = b cos c + c cos B
Using cosine formula

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.9 Additional Questions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

You can Download Samacheer Kalvi 11th 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 11th Maths Solutions Chapter 3 Trigonometry Ex 3.8

Question 1.
Find the principal solution and general solutions of the following:

Solution:

Question 2.
Solve the following equations for which solutions lies in the interval 0° < θ < 360°
(i) sin⁴ = sin²x
(ii) 2 cos² x + 1 = – 3cos x
(iii) 2 sin² x + 1 = 3 sin x
(iv) cos 2x = 1 – 3 sin x – 3 sin x
Solution:

Question 3.
Solve the following equations:
(i) sin 5x – sin x = cos 3x
(ii) 2 cos² θ + 3 sin θ – 3 = 0
(iii) cos θ + cos 3θ = 2 cos 2θ
(iv) sin θ + sin 3θ + sin 5θ = 0
(v) sin 2θ – cos 2θ – sin θ + cos θ = 0
(vi) sin θ + cos θ = \(\sqrt{3}\)
(vii) sin θ + \(\sqrt{3}\) cos θ = 1
(viii) cot θ + cosec θ = \(\sqrt{3}\)

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.8 Additional Questions

Question 1.
Solve: 2 cos² θ + 3 sin θ = 0
Solution:
2 cos² θ + 3 sin θ = 0
⇒ 2 (1 – sin² θ) + 3 sin θ = 0
⇒ 2 – 2 sin² θ + 3 sin θ = 0
⇒ -2 sin² θ + 3 sin θ + 2 = 0
⇒ 2 sin² θ – 3 sin θ – 2 = 0
⇒ (2 sin θ + 1)(sin θ – 2) = 0

Question 2.
Solve: 2 tan θ – cot θ = -1
Solution:
2 tan θ – cot θ = -1

Question 3.
Solve: tan² θ + (1 – \(\sqrt{3}\)) = 0
Solution:

Question 4.
Solve: \(\sqrt{3}\) x + cos x = 2
Solution:

Question 5.

Solution:

You can Download Samacheer Kalvi 12th 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 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.4

Question 1.
Solve the following systems of linear equations by Cramer’s rule:
(i) 5x – 2y + 16 = 0, x + 3y – 7 = 0
(ii) \(\frac{3}{x}+2 y=12, \frac{2}{x}+3 y=13\)
(iii) 3x + 3y – z = 11, 2x – y + 2z = 9, 4x + 3y + 2z = 25

Solution:
(i) 5x – 2y + 16 = 0, x + 3y – 7 = 0
The above equations are 5x – 2y = -16 and x + 3y = -7
The matrix form of two above equations is

Question 2.
In a competitive examination, one mark is awarded for every correct answer while \(\frac { 1 }{ 4 }\) mark is deducted for every wrong answer. A student answered 100 questions and got 80 marks. How many questions did he answer correctly? (Use Cramer’s rule to solve the problem).
Solution:
No. of Questions answered = 100
Let the No. of questions answered correctly be x and the No. of questions answered wrongly be y
Here, x + y = 100 and x – \(\frac { 1 }{ 4 }\) y = 80
(i.e) x + y = 100 and 4x – y = 320
The matrix form is

No. of questions answered correctly x = 84.

Question 3.
A chemist has one solution which is 50% acid and another solution which is 25% acid. How much each should be mixed to make 10 litres of a 40% acid solution? (Use Cramer’s rule to solve the problem).
Solution:
Let the no. of litres in 50% acid used be x litres and the no. of litres in 25% acid used be y litres


(i.e) we have to mix 6 litres in 50% acid and 4 litres in 25% acid.

Question 4.
A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself? (Use Cramer’s rule to solve the problem).
Solution:
Time is taken for pump A to fill the tank be x minutes and time taken for pump B to fill the tank be y minutes


(i.e) Pump A can fill the tank in 15 minutes and pump B can fill the tank in 30 minutes.

Question 5.
A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is ₹ 150. The cost of the two dosai, two idlies and four vadais is ₹ 200. The cost of five dosai, four idlies and two vadais is ₹ 250. The family has ₹ 350 in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had?
Solution:
Let the cost of 1 dosai be ₹ x the cost of 1 idli be ₹ y and the cost of 1 vadai be ₹ z
Here 2x + 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
Writing the above equations in matrix form


So, cost of 1 dosai = x = ₹ 30
cost of 1 idli = y = ₹ 10 and
cost of 1 vadai = z = ₹ 30
Now cost of 3 dosai and 6 idlis and 6 vadais = 3 × 30 + 6 × 10 + 6 × 30 = 90+ 60+ 180 = ₹ 330
and they are having ₹ 350. So they will be able to manage to pay the bill.

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.4 Additional Problems

Question 1.
Solve the following non-homogeneous system of linear equations by determinant method: 3x + 2y = 5, x + 3y = 4.
Solution:

Solution is (x, y) = (1, 1).

Question 2.
Solve the following non-homogeneous system of linear equations by determinant method: x + y + z = 4 ; x – y + z = 2 ; 2x + y – z = 1
Solution:

Since ∆ ≠ 0, the system has unique solution

Solution is x, y, z = 1, 1, 2

Question 3.
Solve the following non-homogeneous system of linear equations by determinant method: 3x + y – z = 2; 2x – y + 2z = 6; 2x + y – 2z = -2
Solution:


Solution is x, y, z = 1, 2, 3

Question 4.
Solve the following non-homogeneous system of linear equations by determinant method:

Solution:

a + 2b – c = 1 … (1)
2a + 4b + c = 5 ….. (2)
3a – 2b – 2c = 0 …… (3)

∴ The system has unique solution.

∴ Solution is x, y, z = 1, 2, 1.