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1: Real Numbers
Algebra
2: Polynomials
3: Linear Equations in Two Variables
4: Quadratic Equations
5: Arithmetic Progression
Coordinate Geometry
6: Coordinate Geometry
Geometry
7: Triangles
8: Circles
9: Constructions
Trigonometry
10: Trignometric Ratios
▶ 11: T-Ratios of Some Particular Angles
12: Trigonometric Ratios of Some Complemantary Angles
13: Trigonometric identities
14: Heights and Distances
Mensuration
15: Perimeter And Area of Plane Figures
16: Area of Circle, Sector and Segment
17: Volumes and Surface Areas of Solids
Statistics and Probability
18: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
19: Probability
Chapter 20: Additional Questions
![R.S. Aggarwal solutions for माठेमटिक्स [इंग्रजी] इयत्ता १० chapter 11 - T-Ratios of Some Particular Angles R.S. Aggarwal solutions for माठेमटिक्स [इंग्रजी] इयत्ता १० chapter 11 - T-Ratios of Some Particular Angles - Shaalaa.com](/images/mathematics-english-class-10_6:8f062ea57bdf49abb4f6d22550b39d56.jpg)
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Solutions for Chapter 11: T-Ratios of Some Particular Angles
Below listed, you can find solutions for Chapter 11 of CBSE, Karnataka Board R.S. Aggarwal for माठेमटिक्स [इंग्रजी] इयत्ता १०.
R.S. Aggarwal solutions for माठेमटिक्स [इंग्रजी] इयत्ता १० 11 T-Ratios of Some Particular Angles EXERCISE 11 [Pages 572 - 574]
Evaluate the following:
sin 60° cos 30° + cos 60° sin 30°
Evaluate the following:
sin 60° cos 30° – cos 60° sin 30°
Evaluate the following:
cos 60° cos 30° – sin 60° sin 30°
Evaluate the following:
cos 45° cos 30° + sin 45° sin 30°
Evaluate the following:
tan 30° cosec 60° + tan 60° sec 30°
Evaluate the following:
2cos260° + 3sin245° – 3sin230° + 2cos290°
Evaluate the following:
`sin^2 30^circ cos^2 45^circ + 4tan^2 30^circ + 1/2 sin^2 90^circ + 1/8 cot^2 60^circ`
Evaluate the following:
`cot^2 30^circ - 2cos^2 30^circ - 3/4 sec^2 45^circ + 1/4 "cosec"^2 30^circ`
Evaluate the following:
(cos 0° + sin 45° + sin 30°)(sin 90° – cos 45° + cos 60°)
Show that `(1 - sin 60^circ)/(cos 60^circ) = (tan 60^circ - 1)/(tan 60^circ + 1)`.
Show that `(cos 30^circ + sin 60^circ)/(1 + sin 30^circ + cos 60^circ) = cos 30^circ`.
Verify the following:
sin 60° cos 30° – cos 60° sin 30° = sin 30°
Verify the following:
cos 60° cos 30° + sin 60° sin 30° = cos 30°
Verify the following:
2 sin 30° cos 30° = sin 60°
Verify the following:
2 sin 45° cos 45° = sin 90°
If A = 45°, verify that sin 2A = 2 sin A cos A.
If A = 45°, verify that cos 2A = 2 cos2A – 1 = 1 – 2 sin2A.
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2A)`.
If A = 30°, verify that `cos 2A = (1 - tan^2A)/(1 + tan^2A)`.
If A = 30°, verify that `tan 2A = (2 tan A)/(1 - tan^2A)`.
If A = 60° and B = 30°, verify that sin (A + B) = sin A cos B + cos A sin B.
If A = 60° and B = 30°, verify that cos (A + B) = cos A cos B – sin A sin B.
If A = 60° and B = 30°, verify that sin (A – B) = sin A cos B – cos A sin B.
If A = 60° and B = 30°, verify that cos (A – B) = cos A cos B + sin A sin B.
If A = 60° and B = 30°, verify that `tan (A - B) = (tan A - tan B)/(1 + tan A tan B)`.
If A and B are acute angles such that `tan A = 1/3, tan B = 1/2` and `tan (A + B) = (tan A + tan B)/(1 - tan A tan B)`, show that A + B = 45°.
Using the formula, `tan 2A = (2 tan A )/(1 - tan^2 A)`, find the value of tan 60°, it being given that `tan 30^circ = 1/sqrt(3)`.
Using the formula, `cos A = sqrt((1 + cos2A)/2)`, find the value of cos 30°, it being given that `cos 60^circ = 1/2`.
Using the formula, `sin A = sqrt((1 - cos 2A)/2)`, find the value of sin 30°, it being given that `cos 60^circ = 1/2`.
If sin (A + B) = sin A cos B + cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of (i) sin 75° (ii) cos 15°.
HINT: Take A = 45° and B = 30°.
If tan (x + 30°) = 1 then find the value of x.
If `(cot θ - 1)/(cot θ + 1) = (1 - sqrt(3))/(1 + sqrt(3))` then find the acute angle θ.
If sin (A + B) = 1 and `tan (A - B) = 1/sqrt(3)`, 0° < (A + B) ≤ 90° and A > B then find the values of A and B.
If `sin (A + B) = sqrt(3)/2` and `cos (A - B) = sqrt(3)/2`, 0° < (A + B) ≤ 90° and A > B then find the values of A and B.
If `tan (A - B) = 1/sqrt(3)` and `tan (A + B) = sqrt(3)`, 0° < (A + B) ≤ 90° and A > B then find the values of A and B.
If cosec (A + B) = 1 and cosec (A – B) = 2, 0° < (A + B) ≤ 90° and A > B then find the values of:
(i) sin A cos B + cos A sin B
(ii) `(tan A - tan B)/(1 + tan A tan B)`
If 3x = cosec θ and `3/x = cot θ`, find the value of `3(x^2 - 1/x^2)`.
In the adjoining figure, ΔABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm. Find (i) BC, (ii) AB.

In the adjoining figure, ΔABC is right-angled at B and ∠A = 30°. If BC = 6 cm, find (i) AB, (ii) AC.

In the adjoining figure, ΔABC is right-angled at B and ∠A = 45°. If AC = `3sqrt(2)` cm, find (i) BC, (ii) AB.

Find the value of x for which x tan 45° cot 60° = sin 30° cosec 60°.
Find the value of x for which `2 "cosec"^2 30^circ + x sin^2 60^circ - 3/4 tan^2 30^circ = 10`.
R.S. Aggarwal solutions for माठेमटिक्स [इंग्रजी] इयत्ता १० 11 T-Ratios of Some Particular Angles MULTIPLE-CHOICE QUESTIONS (MCQ) [Pages 576 - 577]
Choose the correct answer in each of the following questions:
(sec260° – 1) = ?
0
2
3
4
(sin230° – sec260° + 4cot245°) = ?
4
2
1
`1/4`
(3 cos260° + 2 cot230° – 5 sin245°) = ?
1
4
`17/4`
`13/6`
`(cos^2 30^circ cos^2 45^circ + 4 sec^2 60^circ + 1/2 cos^2 90^circ - 2 tan^2 60^circ)` = ?
`81/8`
`83/8`
`73/8`
`75/8`
(cos 0° + sin 30° + sin 45°)(sin 90° + cos 60° – cos 45°) = ?
`7/4`
`5/6`
`3/5`
`5/8`
If tan245° – cos230° = x sin 45° cos 45° then x = ?
2
–2
`1/2`
`-1/2`
If `sqrt(2) sin (60^circ - α) = 1` then α = ?
15°
30°
45°
60°
If tan x = 3 cot x then x = ?
60°
45°
30°
15°
If `sqrt(3) tan 2θ - 3 = 0` then θ = ?
15°
30°
45°
60°
If `2 sin 2θ = sqrt(3)` then θ = ?
30°
45°
60°
90°
If 2 cos 3θ = 1 then θ = ?
10°
15°
20°
30°
If x tan 45° cos 60° = sin 60° cot 60° then x = ?
1
`1/2`
`1/sqrt(2)`
`sqrt(3)`
If tan (3x + 30°) = 1 then x = ?
20°
15°
10°
5°
If sin θ = cos θ, 0 ≤ θ ≤ 90° then θ = ?
30°
45°
60°
90°
If `sin (A - B) = 1/2` and `cos (A + B) = 1/2`, 0° < (A + B) ≤ 90° and A > B then values of A and B are ______.
(60°, 30°)
(60°, 15°)
(45°, 15°)
(60°, 25°)
Solutions for 11: T-Ratios of Some Particular Angles
![R.S. Aggarwal solutions for माठेमटिक्स [इंग्रजी] इयत्ता १० chapter 11 - T-Ratios of Some Particular Angles R.S. Aggarwal solutions for माठेमटिक्स [इंग्रजी] इयत्ता १० chapter 11 - T-Ratios of Some Particular Angles - Shaalaa.com](/images/mathematics-english-class-10_6:8f062ea57bdf49abb4f6d22550b39d56.jpg)
R.S. Aggarwal solutions for माठेमटिक्स [इंग्रजी] इयत्ता १० chapter 11 - T-Ratios of Some Particular Angles
Shaalaa.com has the CBSE, Karnataka Board Mathematics माठेमटिक्स [इंग्रजी] इयत्ता १० CBSE, Karnataka Board solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. R.S. Aggarwal solutions for Mathematics माठेमटिक्स [इंग्रजी] इयत्ता १० CBSE, Karnataka Board 11 (T-Ratios of Some Particular Angles) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.
Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. R.S. Aggarwal textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.
Concepts covered in माठेमटिक्स [इंग्रजी] इयत्ता १० chapter 11 T-Ratios of Some Particular Angles are .
Using R.S. Aggarwal माठेमटिक्स [इंग्रजी] इयत्ता १० solutions T-Ratios of Some Particular Angles exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in R.S. Aggarwal Solutions are essential questions that can be asked in the final exam. Maximum CBSE, Karnataka Board माठेमटिक्स [इंग्रजी] इयत्ता १० students prefer R.S. Aggarwal Textbook Solutions to score more in exams.
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