Test: Complementary Angles of Trigonometry - Class 10 MCQ

Test: Complementary Angles of Trigonometry - Class 10 MCQ

Test Description

15 Questions MCQ Test NTSE for Class 10 - Test: Complementary Angles of Trigonometry

Test: Complementary Angles of Trigonometry for Class 10 2024 is part of NTSE for Class 10 preparation. The Test: Complementary Angles of Trigonometry questions and answers have been prepared according to the Class 10 exam syllabus.The Test: Complementary Angles of Trigonometry MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Complementary Angles of Trigonometry below.
Solutions of Test: Complementary Angles of Trigonometry questions in English are available as part of our NTSE for Class 10 for Class 10 & Test: Complementary Angles of Trigonometry solutions in Hindi for NTSE for Class 10 course. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free. Attempt Test: Complementary Angles of Trigonometry | 15 questions in 15 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study NTSE for Class 10 for Class 10 Exam | Download free PDF with solutions
Test: Complementary Angles of Trigonometry - Question 1

If tan 2A = cot (A – 18°), then the value of A is

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 1

Test: Complementary Angles of Trigonometry - Question 2

The value of cos2 17° – sin2 73° is

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 2

cos217-sin273
=cos217-sin2(90-17)
=cos217-cos217   (because sin(90-x)=cos x)
=0

 1 Crore+ students have signed up on EduRev. Have you?
Test: Complementary Angles of Trigonometry - Question 3

If sec 4A = cosec (A-20°),where 4A is an acute angle, find the value of A

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 3

Test: Complementary Angles of Trigonometry - Question 4

​=

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 4

Taking LCM

Test: Complementary Angles of Trigonometry - Question 5

The value of cos θ cos(90° - θ) – sin θ sin (90° - θ) is:

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 5

Explanation:

- Given expression: cos θ cos(90° - θ) – sin θ sin (90° - θ)
- We know that cos(90° - θ) = sin θ and sin(90° - θ) = cos θ
- Substitute these values into the expression:
= cos θ * sin θ - sin θ * cos θ
= sin θ cos θ - sin θ cos θ
= 0
- Therefore, the value of the expression is 0.

Test: Complementary Angles of Trigonometry - Question 6

Out of the following options, the two angles that are together classified as complementary angles are

Test: Complementary Angles of Trigonometry - Question 7

The value of    is

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 7

we know sin(90 - a) = cos(a)

cos(90 - a) = sin(a)

sin(a) = 1/cosec(a)

sec(a) = 1/cos(a)

cos40 = cos(90-50) = sin50

cosec40 = cosec(90-50) = sec50

so our expression becomes

sin50/sin50 + sec50/sec50 - 4cos50 / sin40

= 1 + 1 - 4(1)   since cos50 = sin40

= -2

Test: Complementary Angles of Trigonometry - Question 8

If cos (40° + A) = sin 30°, the value of A is:​

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 8

cos(θ)=sin(90-θ)
so 40+A+30=90
A=20

Test: Complementary Angles of Trigonometry - Question 9

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 9

Taking LCM

Test: Complementary Angles of Trigonometry - Question 10

The value of   is

Test: Complementary Angles of Trigonometry - Question 11

sin (60° + θ) – cos (30° – θ) is equal to (where (60° + θ) and (30° - θ) are both acute angles):

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 11

Test: Complementary Angles of Trigonometry - Question 12

The value of tan1°.tan2°.tan3°………. tan89° is :

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 12

tan 1.tan 2.tan 3...tan (90 - 3 ).tan ( 90 - 2 ).tan ( 90 - 1)
=tan 1.tan 2 .tan 3...cot 3.cot 2.cot 1
=tan 1.cot 1.tan 2.cot 2.tan 3.cot 3 ... tan 89.cot 89
1 x 1 x 1 x 1 x ... x 1 =1

Test: Complementary Angles of Trigonometry - Question 13

Using the ratio of complementary angles, the value of   is

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 13

Explanation:

Ratio of Complementary Angles:

- Complementary angles are two angles that add up to 90 degrees.
- The ratio of complementary angles is the ratio of the measures of the two angles that add up to 90 degrees.

Given:
Let the two complementary angles be x and y.

Therefore:
x + y = 90

Ratio of Complementary Angles:
The ratio of the two angles can be represented as:
x:y

Using the given information:
x + y = 90
x = 90 - y

Substitute x in the ratio:
(90 - y) : y

Simplify the ratio:
90 : y

Therefore:
The ratio of the two complementary angles is 90 : y.

Conclusion:
Since the ratio of complementary angles is 90 : y, the value of y is 1.

C: 1

Test: Complementary Angles of Trigonometry - Question 14

If x sin (90° – θ) cot (90°- θ) = cos (90° – θ) then x is

Test: Complementary Angles of Trigonometry - Question 15

If A and B are the angles of a right angled triangle ABC, right angled at C, then 1+cot2A =​

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 15
ABC is a Δ, right angle at c.
1 +cot^2 =?........
we know that.....
Cosec^2 - cot^2= 1...
So,
=> 1+ cot^2
=> cosec^2 A
=> (AB)^2/( CB)^2
= sec ^2B.

NTSE for Class 10

338 docs|162 tests
Information about Test: Complementary Angles of Trigonometry Page
In this test you can find the Exam questions for Test: Complementary Angles of Trigonometry solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Complementary Angles of Trigonometry, EduRev gives you an ample number of Online tests for practice

NTSE for Class 10

338 docs|162 tests