Category: Class 10 Maths Objective Questions

[CLASS 10] Mathematics MCQs on Trigonometric Ratios

Mathematics Multiple Choice Questions & Answers on “Trigonometric Ratios – 1”.

1. If sin (A + B) = (frac {sqrt {3}}{2}) and tan (A – B) = 1. What are the values of A and B?
a) 37, 54
b) 35.7, 40.7
c) 50, 10
d) 52.5, 7.5
Answer: d
Clarification: The value of sin (A + B) = (frac {sqrt {3}}{2}) and sin 60° = (frac {sqrt {3}}{2})
∴ A + B = 60     (1)
The value of tan (A – B) = 1 and tan 45° = 1
∴ A – B = 45     (2)
Adding equation (1) and (2)
A + B = 60
+ A – B = 45
– – – – – – – – – – – – –
2 A = 105
A = 52.5
∴ B = 7.5

2. If cos θ = (frac {3}{4}) then value of cos 2θ is ___________
a) (frac {1}{6})
b) (frac {1}{4})
c) (frac {1}{8})
d) (frac {3}{8})
Answer: c
Clarification: cos 2θ = 2cos θ2 – 1
cos θ = (frac {3}{4})
cos 2θ = 2((frac {3}{4}))2 – 1
= (frac {1}{8})

3. If sin A = (frac {8}{17}), what will be the value of cos A sec A?
a) 2
b) -1
c) 1
d) 0
Answer: c
Clarification: sin A = (frac {8}{17})
cos A sec A can be written as cos⁡A × (frac {1}{sec⁡A}) = 1
∴ cos A sec A = 1

4. The value of each of the trigonometric ratios of an angle depends on the size of the triangle and does not depend on the angle.
a) True
b) False
Answer: b
Clarification:
   
Consider, two triangles ABC and DEF
In ∆ABC,
sin B = (frac {AC}{AB} = frac {10}{20} = frac {1}{2}) i.e. B = 30°
Now, in ∆DEF,
sin F = (frac {DE}{DF} = frac {20}{40} = frac {1}{2}) i.e. F = 30°
From these examples it is evident that the value of the trigonometric ratios depends on their angle and not on their lengths.

5. If tan α = √3 and cosec β = 1, then the value of α – β?
a) -30°
b) 30°
c) 90°
d) 60°
Answer: a
Clarification: tan α = √3 and tan 60° = √3
∴ α = 60°
Cosec β = 1 and cosec 90° = 1
∴ β = 90°
α – β = 60 – 90 = -30°

6. In triangle ABC, right angled at C, then the value of cosec (A + B) is __________
a) 2
b) 0
c) 1
d) ∞
Answer: c
Clarification:
Since the triangle is right angles at C,
The sum of the remaining two angles will be 90
∴ cosec(A + B) = Cosec 90° = 1

7. If tan θ = (frac {3}{4}) then the value of sinθ is _________
a) (frac {3}{5})
b) (frac {4}{4})
c) (frac {3}{4})
d) (frac {-3}{5})
Answer: a
Clarification:

tanθ = (frac {BC}{AC} = frac {3}{4} = frac {3k}{4k})
Hence, BC = 3k, AC = 4k
Using Pythagoras theorem
AB2 = AC2 + BC2
AB2 = 4k2 + 3k2
AB = 5k
sinθ = (frac {BC}{AB} = frac {3k}{5k} = frac {3}{5})

8. What is the value of sin30°cos15° + cos30°sin15°?
a) (frac {1}{2})
b) 0
c) 1
d) (frac {1}{sqrt {2}})
Answer: d
Clarification:
sin30°cos15° + cos30°sin15° = sin⁡45° = (frac {1}{sqrt {2}})

9. What is the value of cos A sec A + sin A cosec A – tan A cot A?
a) 0
b) 2
c) 1
d) 3
Answer: c
Clarification:
Cos A sec A = 1
Similarly sin A cosec A = 1 and tan A cot A = 1
∴ cos A sec A + sin A cosec A – tan A cot A = 1 + 1 – 1 = 1

10. In a right angled triangle, the trigonometric function that is equal to the ratio of the side opposite a given angle to the hypotenuse is called cosine.
a) False
b) True
Answer: a
Clarification:

In this ∆ ABC,
sin = (frac {Opposite}{hypotenuse})

[CLASS 10] Mathematics MCQs on Zeros and Coefficients of Polynomial

Mathematics Quiz for Class 10 on “Zeros and Coefficients of Polynomial – 2”.

1. What will be the value of (α – β)2, if α and β are the zeros of 4x2 – 27x – 40?
a) (frac {1369}{16})
b) (frac {139}{16})
c) (frac {1369}{6})
d) (frac {19}{16})
Answer: a
Clarification: α and β are the zeros of 4x2 – 27x – 40
α + β = (frac {-27}{4}) and αβ = – 10
(α – β)2 = α2 + β2 – 2αβ
(α – β)2 = (α + β)2 – 2αβ – 2αβ
(α – β)2 = ((frac {-27}{4}))2 – 4(-10)
(α – β)2 = (frac {729}{16}) + 40 = (frac {1369}{16})

2. If β and γ are the zeros of the polynomial ax3 + bx2 + cx + d, then the value of α2 + β2 + γ2 is _________
a) (frac {b^2-2c}{a^2})
b) (frac {b^2-2ca}{a})
c) (frac {b^2-2ca}{a^2})
d) (frac {b^2+2ca}{a^2})
Answer: c
Clarification: β and γ are the zeros of the polynomial ax3 + bx2 + cx + d
So, α + β + γ = (frac {-b}{a})
αβ + βγ + γα = (frac {c}{a})
Now, α2 + β2 + γ2 = (α + β + γ)2 – 2(αβ + βγ + γα)
α2 + β2 + γ2 = ((frac {-b}{a}))2 – 2((frac {c}{a}) = frac {b^2}{a^2}) – 2 (frac {c}{a} = frac {b^2-2ca}{a^2})

3. The value of αβ + βγ + γα, if α, β and γ are the zeros of 2x3 – 4x2 + 9x – 7 is _______
a) (frac {9}{2})
b) (frac {3}{2})
c) (frac {9}{8})
d) (frac {9}{5})
Answer: a
Clarification: β and γ are the zeros of 2x3 – 4x2 + 9x – 7
The sum of product of two zeros or αβ + βγ + γα = (frac {coefficient , of , x}{coefficient , of , x^3} = frac {9}{2})

4. If α, β and γ are the zeros of 5x3 + 10x2 – x + 20, then the value of αβγ is _______
a) -1
b) 5
c) -10
d) -4
Answer: d
Clarification: α, β and γ are the zeros of 5x3 + 10x2 – x + 20
The product of zeros or αβγ = (frac {-constant , term}{coefficient , of , x^3} = frac {-20}{5}) = -4

5. What will be the polynomial if the value of α + β + γ = -√3 , αβ + βγ + γα = 4 and = (frac {-5}{3})?
a) x3 + 3x2 + 12x + 5
b) 3x3 + √3 x2 + 4x + 5
c) 3x3 + 3√3 x2 + 12x + 5
d) x3 + √3 x2 + 12x + 5
Answer: c
Clarification: α + β + γ = – √3, + βγ + γα = 4, αβγ = (frac {-5}{3})
∴ f(x) = x3 – (α + β + γ) x2 + (αβ + βγ + γα)x – αβγ
Substituting we get,
f(x) = x3 + √3 x2 + 4x + (frac {5}{3})
f(x) = 3x3 + 3√3 x2 + 12x + 5

6. If α, β and γ are the zeros of 2x3 – 6x2 + 5x + 2, then the value of α + β + γ is _______
a) 0
b) 1
c) 3
d) 2
Answer: c
Clarification: α, β and γ are the zeros of 2x3 – 6x2 + 5x + 2
The sum of zeros or α + β + γ = (frac {-coefficient , of , x^2}{coefficient , of , x^3} = frac {6}{2}) = 3

7. If α and β are the zeros of x2 – (5 + 7k)x + 35k, such that α2 + β2 = 172 is then the value of k is _______
a) 6√3
b) √3
c) 3
d) 3√3
Answer: b
Clarification: α and β are the zeros of x2 – (5 + 7k)x + 35k
So, α + β = (5 + 7k) and αβ = 35k
Also, α2 + β2 = 172
(α + β)2 – 2αβ = 172
Substituting values we get,
(5 + 7k)2 – 2(35k) = 172
25 + 49k2 + 70k – 70k = 172
49k2 – 147 = 0
Solving we get, k = √3

8. If α and β are the zeros of x2 + (k2 – 1)x – 20, such that α2 – β2 – αβ = 29 and α – β = 9 then, the value of k is _______
a) 1
b) 0
c) 2
d) 3
Answer: b
Clarification: α and β are the zeros of x2 + (k2 – 1)x – 20
So, α + β = -(k2 – 1) and αβ = -20
Also, α – β = 9
Now, α2 – β2 – αβ = -11
(α + β)(α – β) – αβ = 29
-(k2 – 1)9 + 20 = 29
-(k2 – 1)9 = 9
-(k2 – 1) = 1
(k2 – 1) = -1
k2 = -1 + 1
k = 0

9. If α and β are the zeros of x2 + bx + c, then the polynomial having -α, -β as zeros is _______
a) -x2 – bx + c
b) x2 + bx + c
c) x2 – bx + c
d) x2 – bx – c
Answer: c
Clarification: α and β are the zeros of x2 + bx + c
Also, α + β = -b and αβ = c
Now, -α-β = -(-b) = b and -α × -β = c
Hence, the polynomial with -α, -β as its zeros will be x2 – bx + c

10. If the two zeros of the polynomial x3 – 9x2 -x + 9, are 1 and 9, then the third zero is ________
a) 9
b) 1
c) 2
d) -1
Answer: d
Clarification: The given polynomial is x3 – 9x2 – x + 9.
The two zeros of the polynomial are 1 and 9.
We know that, the sum of zeros of the polynomial or α + β + γ = (frac {-coefficient , of , x^2}{coefficient , of , x^3} = frac {9}{1})
1 + 9 + γ = 9
γ = 9 – 10 = -1

To practice Mathematics Quiz for Class 10,

[CLASS 10] Mathematics MCQs on Trigonometric Identities

Mathematics Aptitude Test for Class 10 on “Trigonometric Identities – 2”.

1. If sec θ – tan θ = M then sec θ + tan θ = (frac {1}{M}).
a) False
b) True
Answer: b
Clarification: The appropriate trigonometric identity used here is sec2 θ – tan2 θ = 1.
(sec θ – tan θ) (sec θ + tan θ) = 1
M (sec θ + tan θ) = 1
(sec θ + tan θ) = (frac {1}{M})

2. Find the correct trigonometric identity.
a) cos2 θ = 1 – sin2 θ
b) cos2 θ = 1 + sin2 θ
c) tan2 θ + sec2 θ = 1
d) tan2 θ = sec2 θ + 1
Answer: a
Clarification: The appropriate trigonometric identity used here is sin2 θ + cos2 θ = 1.
cos2 θ = 1 – sin2 θ

3. Evaluate (sec θ – tan θ) (sec θ + tan θ).
a) 0
b) 1
c) 2
d) 3
Answer: b
Clarification: = (sec θ – tan θ) (sec θ + tan θ) = sec2 θ – tan2 θ
= 1
The identity used here is sec2 θ – tan2 θ = 1

4. Evaluate (sqrt {frac {1 – sin⁡ , ⁡A}{1 + sin , ⁡⁡A}}).
a) cos A + tan A
b) cos A – tan A
c) tan A – cot A
d) sec A + tan A
Answer: d
Clarification: (sqrt {frac {1 – sin⁡ , ⁡A}{1 + sin⁡ , ⁡A}} = sqrt {frac {1 – sin⁡ , ⁡A}{1 + sin , ⁡⁡A}} . sqrt {frac {1 – sin , ⁡A}{1 – sin , ⁡⁡A}})
= (frac {sqrt {(1 + sin , ⁡A)^2}}{sqrt {1 – sin^2} , ⁡A} )
= (frac {1 + sin⁡ , ⁡A}{sqrt {cos^2} , ⁡A})     (∵ sin2 A + cos2 A = 1)
= (frac {1 + sin⁡ , ⁡A}{cos⁡ , ⁡A})
= sec A + tan A

5. Evaluate (cosec2 θ – cot2 θ)2 . (cosec θ + cot θ)2.
a) 1
b) 0
c) (cosec2 θ – cot2 θ)2
d) (cosec θ + cot θ)2
Answer: d
Clarification: (cosec2 θ – cot2 θ)2 . (cosec θ + cot θ)2 = 1 (cosec θ + cot θ)2
= (cosec θ + cot θ)2

6. (1 – sin2 A) (1 + tan2 A) equals to _____
a) – Sec2 θ Tan2 θ
b) – Sec2 θ Tan2 θ
c) 1
d) 0
Answer: c
Clarification: (1 – sin2 A) (1 + tan2 A) = cos2 A . sec2 A
= cos2 A . (frac {1}{cos^2 A})
= 1

7. Evaluate (cosec A – 1) (cosec A + 1) (sec2 A – 1).
a) 0
b) 1
c) (frac {4}{3})
d) (frac {3}{4})
Answer: b
Clarification: (cosec A – 1) (cosec A + 1) (sec2 A – 1) = (cosec2 A – 1) (sec2 A – 1)
= cot2A . tan2 A
= (frac {1}{tan^2 A}) . tan2 A
= 1

8. (sin A + cos A)2 is equal to _____
a) 1 + 2sin A cos A
b) 1 – 2sin A cos A
c) 2sin A cos A – 1
d) 2sin A cos A + 1
Answer: a
Clarification: (sin A + cos A)2 = sin2 A + cos2 A + 2sin A cos A
= 1 + 2sin A cos A

9. Evaluate tan2 A + (1 + sec A) (sec A – 1).
a) 3 tan2 A
b) 0
c) 2tan2 A
d) 1
Answer: c
Clarification: tan2 A + (1 + sec A) (sec A – 1) = tan2 A + (sec A + 1) (sec A – 1)
= tan2 A + (sec2 A – 1)
= tan2 A + tan2 A
= 2tan2 A

10. (1 + cosec⁡ θ) (1 – cosec⁡ θ) + cot2⁡ θ is _____
a) Cot ⁡θ
b) 0
c) 1
d) Tan ⁡θ
Answer: b
Clarification: (1 + cosec⁡ θ) (1 – cosec θ) + cot2 ⁡θ = (1 – cosec2 θ) + cot2 θ
= -cot2⁡ θ + cot2⁡ θ     (∵ cosec2 A – cot2 A = 1)
= 0

To practice Mathematics Aptitude Test for Class 10,

[CLASS 10] Mathematics MCQs on Geometry – Area of Triangle

Mathematics Multiple Choice Questions & Answers on “Geometry – Area of Triangle”.

1. What will be the area of a triangle whose vertices are (3, 1), (0, 4) and (5, 9)?
a) 5 units
b) -5 units
c) -15 units
d) 15 units
Answer: c
Clarification: We know that, area of triangle = (frac {1}{2}){x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2 )}
The coordinates of vertices of the triangle are (3, 1), (0, 4) and (5, 9).
The area of triangle = (frac {1}{2}){3(4 – 9) + 0(9 – 1) + 5(1 – 4) } = (frac {1}{2}) {-15 + 0 – 15} = (frac {-30}{2}) = -15 units
Since area of triangle cannot be zero. So area of triangle is 15 units.

2. What will be the area of the quadrilateral ABCD whose vertices are A (8, 6), B (9, 0), C (1, 2) and D (3, 4)?
a) -20 units
b) 26 units
c) 23 units
d) 3 units
Answer: b
Clarification: Area of quadrilateral = Area of ∆ABC + Area of ∆ADC
We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2)}
The coordinates of vertices of the triangle are (8, 6), (9, 0) and (1, 2).
The area of triangle = (frac {1}{2}){8(0 – 2) + 9(2 – 6) + 1(6 – 0)} = (frac {1}{2}) { – 16 – 36 + 6} = (frac {-46}{2}) = – 23 units
The area of triangle cannot be zero. So, Area of ∆ABC = 23 units
We know that, area of triangle = (frac {1}{2}){x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)}
The coordinates of vertices of the triangle are (8, 6), (3, 4) and (1, 2).
The area of triangle = (frac {1}{2}) {8(4 – 2) + 3(2 – 6) + 1(6 – 4)} = (frac {1}{2}) {16 – 12 + 2} = (frac {6}{2}) = 3 units
So, Area of ∆ADC = 3 units
Area of quadrilateral = Area of ∆ABC + Area of ∆ADC = 23 + 3 = 26 units

3. What will be the area of triangle PQR formed by joining the midpoints of the sides of the triangle whose vertices are (3, 3), (9, 9) and (4, 6)?
a) (frac {1}{2}) units
b) (frac {3}{2}) units
c) (frac {3}{5}) units
d) (frac {5}{2}) units
Answer: b
Clarification:

ABC is the triangle formed by the coordinates A (3, 3), B (9, 9) and C (4, 6) and DEF is the triangle formed by joining the midpoints of AC, BC, AB.
D is the midpoint of AC. Coordinates of D = ( ( frac {x_1+x_2}{2}, frac {y_1+y_2}{2} ) = ( frac {3+4}{2}, frac {3+6}{2} ) = ( frac {7}{2}, frac {9}{2} ) )
E is the midpoint of AB. Coordinates of E = ( ( frac {x_1+x_2}{2}, frac {y_1+y_2}{2} ) = ( frac {3+9}{2}, frac {3+9}{2} ) = ( frac {12}{2}, frac {12}{2} ) ) = (6, 6)
F is the midpoint of BC. Coordinates of F = ( ( frac {x_1+x_2}{2}, frac {y_1+y_2}{2} ) = ( frac {9+4}{2}, frac {9+6}{2} ) = ( frac {13}{2}, frac {15}{2} ) )
The area of triangle formed by D ( ( frac {7}{2}, frac {9}{2} ) ), E (6, 6) and F( ( frac {13}{2}, frac {15}{2} ) )
We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}
The coordinates of vertices of the triangle are ( ( frac {7}{2}, frac {9}{2} ) ), (6, 6) and ( ( frac {13}{2}, frac {15}{2} ) ).
The area of triangle = (frac {1}{2} { frac {7}{2}) (6 – (frac {15}{2})) + 6((frac {15}{2} – frac {9}{2})) + (frac {13}{2} ( frac {9}{2} – 6 ) } ) = (frac {1}{2} { – frac {21}{4}) + 18 – (frac {39}{4} } = frac {3}{2} ) units

4. If the points A, B, C is collinear then the area of the triangle will be zero.
a) True
b) False
Answer: a
Clarification: We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}
Let us assume the vertices of the triangle are (0, 0), (0, 4) and (0, 9).
The area of triangle = (frac {1}{2}) {0(4 – 9) + 0(9 – 0) + 0(0 – 4) } = (frac {1}{2}){0} = 0 units
Hence, the area of triangle will be zero if the points are collinear.

5. What will be the area of triangle formed by joining the points P (2, 3), Q (4, 6) and R (6, 9)?
a) 0
b) 2
c) 3
d) 4
Answer: a
Clarification: We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}
Let us assume the vertices of the triangle are (2, 3), (4, 6) and (6, 9).
The area of triangle = (frac {1}{2}) {2(6 – 9) + 4(9 – 3) + 6(3 – 6) } = (frac {1}{2}){0} = 0 units
Hence, the area of triangle is zero, the points are collinear.

6. What will be the value of p, for which the points (-3, p), (-1, 2) and (1, 1) are collinear?
a) p = -4
b) p = 4
c) p = 3
d) p = -3
Answer: c
Clarification: We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}
The vertices of the triangle are (-3, p), (-1, 2) and (1, 1)
The area of triangle = (frac {1}{2}) {-3(2 – 1) + (-1)(1 – p) + 1(p – 2)} = (frac {1}{2}) {-3 – 1 + p + p – 2} = (frac {-6 + 2p}{2}) = – 3 + p
Since, the points are collinear; the area of triangle will be zero.
-3 + p = 0
p = 3

7. The points A (3, 0), B (4, 5) and C (6, 7) are collinear.
a) True
b) False
Answer: b
Clarification: We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )}
Let us assume the vertices of the triangle are A (3, 0), B (4, 5) and C (6, 7)
The area of triangle = (frac {1}{2}) {3(5 – 7) + 4(7 – 0) + 6(0 – 5)} = (frac {1}{2}) {-8} = – 4 units
Since, the area of triangle is not zero.
Therefore, the points are not collinear.

8. Three points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear only when (frac {1}{2}) {x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 ) } = 0.
a) False
b) True
Answer: b
Clarification: Consider three points (-3, 3), (-1, 2) and (1, 1)
We know that, area of triangle = (frac {1}{2}){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2)}
The area of triangle = (frac {1}{2}) {-3(2 – 1) + (-1)(1 – 3) + 1(3 – 2)} = (frac {1}{2}) {-3 – 1 + 3 + 3 – 2} = (frac {0}{2}) = 0
Hence if the points are collinear the area of triangle is zero.

9. If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a triangle ABC, then its area is given by (frac {1}{2}) {x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 ) }
a) True
b) False
Answer: a
Clarification:

A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a triangle ABC.
Draw AL, CM, BN perpendicular to x – axis.
Then, ML = (x1 – x2), LN = (x3 – x1) and MN = (x3 – x2).
∴ area of ∆ABC = ar(trap.BMLA) + ar(trap.ALNC) + ar(trap.BMNC)
= {(frac {1}{2}) (AL + BM) × ML} + {(frac {1}{2}) (AL + CN) × LN} – {(frac {1}{2}) (CN + BM) × MN}
= {(frac {1}{2}) (y1 + y2 ) × (x1 – x2)} + {(frac {1}{2}) (y1 + y3 ) × (x3 – x1)} – {(frac {1}{2}) (y2 + y3 ) × (x3 – x2)}
= (frac {1}{2}) {x1 (y1 + y2 – y1 – y3 ) + x2 (y2 + y3 – y1 – y2 ) – x3 (y1 + y3 – y2 – y3 ) }
= (frac {1}{2}) {x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 ) }

10. If the points A (a, 0), B (0, b) and C (c, 0) are collinear then a = -c.
a) False
b) True
Answer: a
Clarification: We know that, area of triangle = (frac {1}{2}){x1(y2 – y3 ) + x2(y3 – y1 ) + x3(y1 – y2 )}
The three points are A (a, 0), B (0, b) and C (c, 0)
Since the points are collinear, the area of triangle will be zero.
The area of triangle = (frac {1}{2}) {a(b – 0) + 0(c – 0) + c(0 – b) } = (frac {1}{2}) {ab + 0 – bc}
(frac {1}{2}) {ab – bc} = 0
{ab – bc} = 0
a = c

[CLASS 10] Mathematics MCQs on Zeros and Coefficients of Polynomial

Mathematics Multiple Choice Questions & Answers on “Zeros and Coefficients of Polynomial – 1”.

1. The zeros of the polynomial 18x2-27x+7 are ___________
a) (frac {7}{6}, frac {1}{3})
b) (frac {-7}{6}, frac {1}{3})
c) (frac {7}{6}, frac {-1}{3})
d) (frac {7}{3}, frac {1}{3})
Answer: a
Clarification: 18x2-27x+7=0
18x2-21x-6x+7=0
3x(6x-7)-1(6x-7)=0
(6x-7)(3x-1)=0
x=(frac {7}{6}, frac {1}{3})
The zeros are (frac {7}{6}) and (frac {1}{3}).

2. What will be the polynomial if its zeros are 3, -3, 9 and -9?
a) x4-80x2+729
b) x4-90x2+729
c) x4-90x2+79
d) x4-100x2+729
Answer: b
Clarification: The zeros of the polynomial are 3, -3, 9 and -9.
Then, (x-3), (x+3), (x-9) and (x+9) are the factors of the polynomial.
Multiplying the factors, we have
(x-3) (x+3) (x-9) (x+9)
(x2-9) (x2-81) (By identity (x-a)(x+a)=x2-a2)
(x4-9x2-81x2+729)
x4-90x2+729

3. The sum and product of zeros of a quadratic polynomial are 10 and (frac {5}{2}) respectively. What will be the quadratic polynomial?
a) 2x2-20x+10
b) 2x2-x+5
c) 2x2-20x+5
d) x2-20x+5
Answer: c
Clarification: The sum of the polynomial is 10, that is, α+β = 10
The product of the polynomial is (frac {5}{2}) i.e. αβ = (frac {5}{2})
∴ f(x)=x2-(α+β)x+αβ
f(x)=x2-10x+(frac {5}{2})
f(x)=2x2-20x+5

4. If α and β are the zeros of x2+20x-80, then the value of α+β is _______
a) -15
b) -5
c) -10
d) -20
Answer: d
Clarification: α and β are the zeros of x2+20x-80.
Sum of zeros or α+β = (frac {-coefficient , of , x}{coefficient , of , x^2} = frac {-20}{1}) = -20

5. If α and β are the zeros of 3x2-5x-15, then the value of αβ is _______
a) -5
b) -10
c) -15
d) -20
Answer: a
Clarification: α and β are the zeros of 3x2-5x-15.
Product of zeros or αβ = (frac {constant , term}{coefficient , of , x^2} = frac {-15}{3}) = -5

6. What will be the value of other zero, if one zero of the quadratic polynomial is 5 and the sum of the zeros is 10?
a) 10
b) 5
c) -5
d) -10
Answer: b
Clarification: One zero of the quadratic polynomial is 5. ∴ the factor of the polynomial is (x-5)
Let us assume the other zero to be b. ∴ the other factor of the polynomial is (x-b)
Multiplying the factors, we have (x-5)(x-b)
x2-5x-bx+5b
x2-(5+b)x+5b
The sum of zeros is 10.
∴ (frac {-coefficient , of , x}{coefficient , of , x^2})=10
(frac {-(-5-b)}{1})=10
5+b=10
b=5
The equation becomes x2-10x+25.
Therefore, the other zero is 5.

7. The value of a and b, if the zeros of x2+(a+5)x-(b-4) are -5 and 9 will be _________
a) 47, -5
b) -5, 47
c) -9, 49
d) -4, 45
Answer: c
Clarification: The zeros of the polynomial are -5 and 9.
Hence, α=-5, β=9
The polynomial is x2+(a+5)x-(b-4).
Sum of zeros or α+β=-5+9 = (frac {-coefficient , of , x}{coefficient , of , x^2} = frac {a+5}{1})
-4=a+5
a = -9
Product of zeros or αβ = -45 = (frac {constant , term}{coefficient , of , x^2} = frac {-(b-4)}{1})
-45=-b+4
b=49

8. What will be the value of k, if one zero of x2+(k-3)x-16=0 is additive inverse of other?
a) 4
b) -4
c) -3
d) 3
Answer: d
Clarification: Since, one zero of the polynomial is the additive inverse of the other.
Hence, the sum of roots will be zero.
The polynomial is x2+(k-3)x-16=0
Sum of zeros or α+β=(frac {-coefficient , of , x}{coefficient , of , x^2} = frac {k-3}{1})=0
k-3=0
k=3

9. If α and β are the zeros of 10x2+20x-80, then the value of (frac {1}{alpha } + frac {1}{beta }) is _______
a) (frac {5}{4})
b) (frac {1}{5})
c) (frac {3}{4})
d) (frac {1}{4})
Answer: d
Clarification: (frac {1}{alpha } + frac {1}{beta } = frac {alpha +beta }{alpha beta })
α+β=(frac {-20}{10})=-2
αβ=(frac {-80}{10})=-8
∴ (frac {alpha +beta }{alpha beta } = frac {-2}{-8} = frac {1}{4})

10. If α and β are the zeros of x2+35x-75, then _______
a) α+β<αβ
b) α+β>αβ
c) α+β=αβ
d) α+β≠αβ
Answer: b
Clarification: The given polynomial is x2+35x-75.
The sum of zeros, α + β = (frac {-coefficient , of , x}{coefficient , of , x^2} = frac {-35}{1}) = -35
The product of zeros, αβ = (frac {constant , term}{coefficient , of , x^2}) = -75
Clearly, sum of zeros is greater than product of zeros.

[CLASS 10] Maths MCQs on Trigonometry Application – Height & Distance

Mathematics Multiple Choice Questions & Answers on “Trigonometry Application – Height and Distance”.

1. The distance between two objects can be known using trigonometric ratios.
a) True
b) False

Answer: a
Clarification: Finding the distance between two objects is one of the different applications of trigonometry. Trigonometry is also used to find the height or length of an object too.

2. One of the applications of trigonometry is used for astronomers to find the distance between two planets.
a) False
b) True

Answer: b
Clarification: The length or height or distance between two objects can be known using trigonometry. Trigonometry is applied in astronomy, navigation, architecture, etc.

3. _____ is drawn from the eye of an observer to the targeted object.
a) A parallel line
b) Line of sight
c) Elevation line
d) Depression line

Answer: b
Clarification: The line of sight is a line that is drawn from the eye of an observer to the targeted object viewed by the observer and the line of sight is required to form the angle of elevation.

4. What happens to the angle of elevation if the height of a tower, the distance between the tower and the observer is doubled?
a) Doubled
b) halved
c) Tripled
d) Remains the same

Answer: d
Clarification: The angle of elevation is independent of the height of an object and the distance between the object and the observer. So, the angle of elevation remains the same when the height of a tower, the distance between the tower and the observer is doubled.

5. The angle of depression is always a/an _____ angle.
a) right
b) obtuse
c) complete
d) acute

Answer: d
Clarification: The angle of elevation is an acute angle. The angles that are less than 90° are called as acute angles. The angle of depression is also an acute angle.

6. A pole stands vertically on the floor. From a point, it is 120 meters away from the foot of the pole and the angle of elevation is 45° then find the height of the pole.
a) 100 meters
b) 180 meters
c) 120 meters
d) 10 meters
Answer: c

7. A slope is built against a wall which makes an angle 30° with the ground and the height of the wall is 2 meters. Find the length of the slope in meters.
a) 2
b) 4
c) 1
d) 3
Answer: b

8. A slide should be built in a park that has to be set up at 10 meters height and also 10 meters away from the support. What should be the angle of elevation?
a) 45°
b) 30°
c) 60°
d) 90°
Answer: a

9. A tree trunk of height 12 meters is erected with the support of three metal chains. An angle of 60° is made by each metal chain with the tree trunk. Find the length of the metal chain.
a) 32 meters
b) 10meters
c) 24 meters
d) 18 meters
Answer: c

10. A ship has to cross a sea by making an angle of 30° and with a distance of 200 meters to reach the shore then find the width of the sea.
a) 300 meters
b) 100 meters
c) 110 meters
d) 50 meters
Answer: b