Loading [MathJax]/jax/element/mml/optable/Latin1Supplement.js
MCQExams
0:0:2
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 1 - MCQExams.com
CBSE
Class 10 Maths
Introduction To Trigonometry
Quiz 1
The value of
2
t
a
n
30
∘
is ________
Report Question
0%
3.46
0%
3.56
0%
34.6
0%
3.48
Explanation
2
tan
30
∘
=
2
1
√
3
=
2
√
3
=
3.46
The value of
tan
θ
×
cot
θ
=
_____
Report Question
0%
2
0%
0
0%
1
0%
−
1
Explanation
To find: Value of
tan
θ
×
cot
θ
We know that
cot
θ
=
1
tan
θ
So,
tan
θ
×
1
tan
θ
=
1
Which of the following is equal to
sin
x
sec
x
?
Report Question
0%
tan
x
0%
cot
x
0%
cos
x
tan
x
0%
cos
x
cosec
x
0%
cot
x
cosec
x
Explanation
sin
x
⋅
sec
x
=
sin
x
×
1
cos
x
=
sin
x
cos
x
=
tan
x
The value of
sin
θ
×
cosec
θ
=
_____
Report Question
0%
2
0%
1
0%
−
1
0%
0
Explanation
sin
θ
×
c
o
s
e
c
θ
=
sin
θ
×
1
sin
θ
=
1
Hence, the answer is
1.
Who published the trigonometry in 1595?
Report Question
0%
William Rowan Hamilton
0%
Hipparchus
0%
Bartholomaeus Pitiscus
0%
Newton
Explanation
Trigonometry was first published by Bartholomaeus Pitiscus in 1595.
So, option C is correct.
Choose the correct alternative answer for the following question.
c
o
s
e
c
45
∘
=
?
Report Question
0%
1
2
0%
√
2
0%
√
3
2
0%
2
√
3
Explanation
sin
45
∘
=
1
√
2
c
o
s
e
c
45
∘
=
√
2
Hence, the correct answer is
√
2
.
If
√
3
cos
A
=
sin
A
, then the value of
cot
A
is:
Report Question
0%
√
3
0%
1
0%
1
√
3
0%
2
tan
45
∘
=
?
Report Question
0%
1
0%
√
3
0%
0
0%
can't be determined
Explanation
tan
45
∘
=
sin
45
∘
cos
45
∘
=
1
√
2
1
√
2
=
1
So the relation is
True
The value of
cos
1
o
.
cos
2
o
.
cos
3
o
.
.
.
.
.
cos
179
o
is equal to
Report Question
0%
−
1
0%
0
0%
1
0%
1
√
2
Explanation
A
=
cos
1
0
⋅
cos
2
0
⋅
cos
3
0
⋯
cos
90
0
⋯
cos
179
0
∵
cos
90
0
=
0
∴
A
=
0
Hence, option B.
In
△
A
B
C
,
∠
B
=
90
∘
,
sin
C
=
3
5
, then
cos
A
=
______
Report Question
0%
3
5
0%
4
5
0%
5
4
0%
5
3
Explanation
In a right angled triangle,
A
+
B
+
C
=
180
∘
B
=
90
∘
So,
A
+
C
=
90
∘
A
=
90
∘
−
C
cos
A
=
cos
(
90
∘
−
A
)
⇒
cos
A
=
sin
C
=
3
5
⇒
cos
A
=
3
5
If sin
A
=
1
2
and cos
B
=
1
2
, then the value of
(
A
+
B
)
is equal to :
Report Question
0%
0
∘
0%
60
∘
0%
90
∘
0%
30
∘
Explanation
sin
A
=
1
2
,
cos
B
=
1
2
\Rightarrow \sin { A } =\sin { 30° } ,\cos { B } =\cos { 60° }
\Rightarrow A=30°,B=60°
\therefore A+B=30°+60°=90°
Hence, the answer is
90°.
If cot
\theta =\dfrac{7}{8}
, then the value of
tan^{2}\theta
equals to :
Report Question
0%
\dfrac{8}{7}
0%
\dfrac{49}{64}
0%
\dfrac{64}{49}
0%
\dfrac{7}{8}
Explanation
\cot { \theta } =\dfrac { 7 }{ 8 }
\Rightarrow \tan { \theta } =\dfrac { 1 }{ \cot { \theta } } \\\ \ \ \ \ \ \ \ \ \ \ \ =\dfrac { 8 }{ 7 }
\Rightarrow \tan ^{ 2 }{ \theta } =\dfrac { 64 }{ 49 }
Hence, the correct option is (C)
The value of (sin
45^{\circ}+cos 45^{\circ}
) is :
Report Question
0%
1
0%
\dfrac{1}{\sqrt{2}}
0%
\dfrac{\sqrt{3}}{2}
0%
\sqrt{2}
Explanation
\sin { 45° } +\cos { 45° }
=\dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ \sqrt { 2 } }
=\dfrac { 2 }{ \sqrt { 2 } }
=\sqrt { 2 }
Hence, the answer is
=\sqrt { 2 } .
If
2 \cos\theta+ \sin\theta=1
, then the value of
4 \cos\theta + 3 \sin\theta
is equal to
Report Question
0%
3
0%
-5
0%
\dfrac {7}{5}
0%
-4
\cos ^{2} 60^{\circ}+\sin ^{2} 30^{\circ} = \displaystyle \frac{1}{a}
Then
a=
Report Question
0%
2
0%
3
0%
1
0%
\dfrac{1}{2}
Explanation
\cos^2 60^{\circ} + \sin^2 30^{\circ}
=\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2
=\cfrac{1}{4} + \cfrac{1}{4}
=\cfrac{2}{4}
=\cfrac{1}{2}
\therefore \dfrac1a = \dfrac12
\Rightarrow a = 2
If
\sec 4A = cosec (A-20^{\small\circ})
, where
4A
is an acute angle, find the value of
A
.
Report Question
0%
A = 32^{\small\circ}
0%
A = 22^{\small\circ}
0%
A = 41^{\small\circ}
0%
A = 16^{\small\circ}
Explanation
Hence
\sin(A-20^{0})=\cos(4A)
\sin(A-20^{0})=\sin(90^{0}-4A)
Hence
A-20^{0}=90^{0}-4A
5A=110^{0}
A=22^{0}
In the given figure,
BC=15cm
and
\sin{B}=\dfrac{4}{5}
, What is the value of
AB
?
Report Question
0%
25cm
0%
20cm
0%
5cm
0%
4cm
Explanation
\sin B= \dfrac{AC}{AB} =\dfrac{\sqrt{AB^2 - BC^2}}{AB}
\dfrac{4}{5} AB= \sqrt{AB^2 - 15^2}
\Rightarrow 16AB^2 = 25AB^2 - (25)(15)^2
AB^2 = \dfrac{25\times 15 \times 15}{9}
\Rightarrow \boxed {AB = 25}
Choose the correct option. Justify your choice.
\displaystyle 9{ \sec }^{ 2 }A-9{ \tan }^{ 2 }A=
Report Question
0%
1
0%
9
0%
8
0%
0
Explanation
We know,
1= \sec ^2A-\tan^2A
\therefore 9(\sec^2A-\tan^2A)
=9\times 1=9
If
sin({ 90 }^{ 0 }-\theta )=\dfrac { 3 }{ 7 }
, then
cos\theta
Report Question
0%
\cfrac { 7 }{ 3 }
0%
\cfrac { 3 }{ 7 }
0%
\cfrac { -7 }{ 3 }
0%
\cfrac { -3 }{ 7 }
Explanation
\sin (90^o - \theta) = \dfrac{3}{7}
We know,
\sin (90^o - \theta) = \cos \theta
Hence,
\cos \theta = \dfrac{3}{7}
If
\tan \theta= \cot \theta
, then the value of
\sec \theta
is :
Report Question
0%
2
0%
1
0%
\dfrac{2}{\sqrt{3}}
0%
\sqrt{2}
Explanation
\tan { \theta } =\cot { \theta }
\Rightarrow \tan ^{ 2 }{ \theta } =1
\Rightarrow \tan { \theta } =1
\Rightarrow \theta =45°
\therefore \sec { 45° } =\sqrt { 2 }
Hence, the answer is
\sqrt { 2 }.
If
\sqrt{3}\,=\,1.732
, find (correct to two decimal places) the value of each of the following :
sin\,60^{\circ}
is 0.87
State true or false.
Report Question
0%
True
0%
False
Explanation
\sin 60^{\circ} = \dfrac{\sqrt{3}}{2} = \dfrac{1.732}{2} = 0.87
The value of cosec
30^{\circ}
+ cot
45^{\circ}
is:
Report Question
0%
2
0%
1
0%
3
0%
-1
Explanation
cosec
30^{\circ}
+ cot
45^{\circ}
=2+1
=3
Hence, the answer is
3.
The value of
5cosec^{2}\theta -5cot^{2 }\theta
is equal to:
Report Question
0%
5
0%
1
0%
0
0%
-5
Explanation
Using the identity:
cosec^{2}\theta -\cot^{2 }\theta=1
We get,
5cosec^{2}\theta -5\cot^{2 }\theta
= 5(
cosec^{2}\theta -\cot^{2 }\theta)=5
If
\theta =45^{\circ},
then the value of
cosec^{2}\theta
is
Report Question
0%
\dfrac{1}{\sqrt{2}}
0%
1
0%
\dfrac{1}{2}
0%
2
Explanation
\theta =45°
\Rightarrow cosec^{ 2 }45°={ \left( \sqrt { 2 } \right) }^{ 2 }=2
Hence, the answer is
2.
For an acute angle
\theta
in a right angled triangle,
\dfrac{1}{\sin{\theta}} =
Report Question
0%
\cos{\theta}
0%
\cot{\theta}
0%
\sec{\theta}
0%
\text{cosec}\,{\theta}
Explanation
\sin \theta=\dfrac{opposite}{hypotenuse}
\cos \theta = \dfrac{adjacent}{hypotenuse}
\dfrac{1}{\sin \theta}=\dfrac{1}{\dfrac{opposite}{hypotenuse}}=\dfrac{hypotenuse}{opposite}=\text{cosec}\ \theta
\therefore\ \dfrac{1}{\sin \theta}=\text{cosec}\ \theta
What is the value of
\displaystyle \sin { { 30 }^{ o } } \times co\sec{ 30 }^{ o }
?
Report Question
0%
1
0%
0
0%
\displaystyle \frac { 1 }{ 2 }
0%
2
Explanation
\displaystyle \sin { { 30 }^{ o }\times co\sec{ 30 }^{ o } } =\cfrac { 1 }{ 2 } \times \cfrac { 2 }{ 1 } =\cfrac { 1 }{ 1 } =1
The value of
\displaystyle \tan { { 45 }^{ o } } \times \cot{ { 45 }^{ o } }
is :
Report Question
0%
0
0%
1
0%
2
0%
\displaystyle \frac { 1 }{ 2 }
Explanation
\displaystyle \tan { { 45 }^{ o } } \times \cot{ { 45 }^{ o } }=1\times 1=1
\dfrac{1}{\cos{\theta}} =
_____ , for an acute angle
\theta
in a right angled triangle.
Report Question
0%
\text{cosec}\,{\theta}
0%
\sec{\theta}
0%
\cot{\theta}
0%
None of these
Explanation
\sin \theta=\dfrac{opposite}{hypotenuse}
\cos \theta = \dfrac{adjacent}{hypotenuse}
\dfrac{1}{\cos \theta}=\dfrac{1}{\dfrac{adjacent}{hypotenuse}}=\dfrac{hypotenuse}{adjacent}=\sec \theta
\therefore\ \dfrac{1}{\cos \theta}=\sec \theta
The value of
\tan{\theta}\cdot\tan{(90-\theta)}
is equal to
Report Question
0%
\sin^2{\theta}
0%
1
0%
\cos^2{\theta}
0%
0
Explanation
\tan\theta \cdot \tan(90-\theta) = \tan\theta \times \cot\theta = 1
\because \tan(90-\theta) =\cot \theta
If
\sec{2A}=\csc{(A-42^\circ)}
where
2A
is acute angle then value of
A
is
Report Question
0%
44^\circ
0%
22^\circ
0%
21^\circ
0%
66^\circ
Explanation
\sec{2A}=\csc{(A-42^\circ)}
\csc{(90-2A)}=\csc{(A-42^\circ)}
90-2A=A-42
3A=132
A=44^\circ
0:0:2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
1
Not Answered
29
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 10 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page