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Shubhangi B. Class 11 Tuition trainer in Sahibzada Ajit Singh Nagar

Shubhangi B.

Achieve excellence in Commerce with expert

Sector 61, Sahibzada Ajit Singh Nagar, India - 160062.

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Referral Discount: Get ₹ 250 off when you make a payment to start classes. Get started by Booking a Demo.

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Overview

Shubhangi B. describes herself as Achieve excellence in Commerce with expert. She conducts classes in BBA Tuition, BCom Tuition and Class 10 Tuition. Shubhangi is located in Sector 61, Sahibzada Ajit Singh Nagar. Shubhangi takes Regular Classes- at her Home. She has 3 years of teaching experience . Shubhangi is pursuing Doctor of Philosophy (Ph.D.) from Panjab University . Shubhangi has completed Bachelor of Commerce (B.Com.) from Panjab University in 2017 and Master of Commerce (M.Com.) from Panjab University in 2019. She is well versed in English.

Languages Spoken

English

Education

Panjab University Pursuing

Doctor of Philosophy (Ph.D.)

Panjab University 2017

Bachelor of Commerce (B.Com.)

Panjab University 2019

Master of Commerce (M.Com.)

National Testing Agency 2019

UGC NET IN COMMERCE

Address

Sector 61, Sahibzada Ajit Singh Nagar, India - 160062

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Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

3

Board

CBSE

CBSE Subjects taught

Accountancy

Taught in School or College

No

Teaching Experience in detail in Class 11 Tuition

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

3

Board

CBSE

CBSE Subjects taught

Accountancy

Taught in School or College

No

Teaching Experience in detail in Class 12 Tuition

https://www.youtube.com/channel/UCmBrjuC7Uu2mRy5op27KOiQ

BCom Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in BCom Tuition

3

BCom Subject

Corporate Accounting, Micro & Macro Economics, Business Mathematics and Statistics, Financial Management, Financial Accounting

Type of class

Regular Classes

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

BBA Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in BBA Tuition

3

BBA Subject

Fundamentals of Accounting, Financial Accounting, Statistics, Corporate Accounting

Type of class

Regular Classes

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

2

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

2

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

IELTS Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 12 do you teach for?

CBSE

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach BBA Tuition, BCom Tuition, Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 9 Tuition and IELTS Coaching Classes.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for 3 years.

Answers by Shubhangi B. (4)

Answered on 01/08/2019 Learn CBSE/Class 10/Mathematics/UNIT V: Trigonometry/Introduction to Trigonometry/NCERT Solutions/Exercise 8.1

Step-by-step explanation: Given i ) = /* From (1) */ = = = =---(1) ii) = After cancellation, we get = ---(2) Therefore, ...more

CosA = \frac{\sqrt{7}}{4}

tanA=\frac{3}{\sqrt{7}}

Step-by-step explanation:

Given sinA = \frac{3}{4}

i ) cosA = \sqrt{1-sin^{2}A}

=\sqrt{1-\left(\frac{3}{4}}\right)^{2}

/* From (1) */

=\sqrt{1-\frac{9}{16}}

=\sqrt{\frac{16-9}{16}}

=\sqrt{\frac{7}{16}}

=\frac{\sqrt{7}}{4}---(1)

ii) tanA = \frac{SinA}{cosA}

= \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}}

After cancellation, we get

= \frac{3}{\sqrt{7}}---(2)

Therefore,

CosA = \frac{\sqrt{7}}{4}

tanA=\frac{3}{\sqrt{7}}

 

Answers 3 Comments
Dislike Bookmark

Answered on 01/08/2019 Learn CBSE/Class 10/Mathematics/UNIT V: Trigonometry/Introduction to Trigonometry/NCERT Solutions/Exercise 8.1

Using pythagoras theorem, (PR) ² = (PQ)² + (PR)² ⇒(QR)² = (13)² - (12) =169 – 144 = 25 QR = 5 cm tan p - cot r= QR/PQ - QR/PQ= 5/12 - 5/12=0 ...more

Using pythagoras theorem,

(PR) ² = (PQ)² + (PR)²

 

 

⇒(QR)² = (13)² - (12)

=169 – 144 = 25

 QR = 5 cm



tan p - cot r


= QR/PQ - QR/PQ

= 5/12 - 5/12

=0

 

Answers 3 Comments
Dislike Bookmark

Answered on 01/08/2019 Learn CBSE/Class 10/Mathematics/UNIT V: Trigonometry/Introduction to Trigonometry/NCERT Solutions/Exercise 8.1

In Δ ABC, right-angled at B Using Pythagoras theorem AC² = AB² +BC² AC² = 576 + 49 = 625 AC = √625 AC = ±25 Now (i) In a right angle triangle ABC where B=90° , Sin A = BC/AC =7/25 CosA =AB/AC =24/25 (ii) Sin C =AB/AC =24/25 Cos C =BC/AC =7/25 ...more

In Δ ABC, right-angled at B

Using Pythagoras theorem

AC² = AB² +BC²  

AC² = 576 + 49 = 625

AC = √625

AC = ±25

 

Now

(i) In a right angle triangle ABC where B=90° ,

Sin A = BC/AC

=7/25

CosA =AB/AC

=24/25

(ii) Sin C =AB/AC

=24/25

Cos C =BC/AC

=7/25

Answers 3 Comments
Dislike Bookmark

Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

3

Board

CBSE

CBSE Subjects taught

Accountancy

Taught in School or College

No

Teaching Experience in detail in Class 11 Tuition

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

3

Board

CBSE

CBSE Subjects taught

Accountancy

Taught in School or College

No

Teaching Experience in detail in Class 12 Tuition

https://www.youtube.com/channel/UCmBrjuC7Uu2mRy5op27KOiQ

BCom Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in BCom Tuition

3

BCom Subject

Corporate Accounting, Micro & Macro Economics, Business Mathematics and Statistics, Financial Management, Financial Accounting

Type of class

Regular Classes

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

BBA Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in BBA Tuition

3

BBA Subject

Fundamentals of Accounting, Financial Accounting, Statistics, Corporate Accounting

Type of class

Regular Classes

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

2

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

2

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

IELTS Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

No Reviews yet!

Answers by Shubhangi B. (4)

Answered on 01/08/2019 Learn CBSE/Class 10/Mathematics/UNIT V: Trigonometry/Introduction to Trigonometry/NCERT Solutions/Exercise 8.1

Step-by-step explanation: Given i ) = /* From (1) */ = = = =---(1) ii) = After cancellation, we get = ---(2) Therefore, ...more

CosA = \frac{\sqrt{7}}{4}

tanA=\frac{3}{\sqrt{7}}

Step-by-step explanation:

Given sinA = \frac{3}{4}

i ) cosA = \sqrt{1-sin^{2}A}

=\sqrt{1-\left(\frac{3}{4}}\right)^{2}

/* From (1) */

=\sqrt{1-\frac{9}{16}}

=\sqrt{\frac{16-9}{16}}

=\sqrt{\frac{7}{16}}

=\frac{\sqrt{7}}{4}---(1)

ii) tanA = \frac{SinA}{cosA}

= \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}}

After cancellation, we get

= \frac{3}{\sqrt{7}}---(2)

Therefore,

CosA = \frac{\sqrt{7}}{4}

tanA=\frac{3}{\sqrt{7}}

 

Answers 3 Comments
Dislike Bookmark

Answered on 01/08/2019 Learn CBSE/Class 10/Mathematics/UNIT V: Trigonometry/Introduction to Trigonometry/NCERT Solutions/Exercise 8.1

Using pythagoras theorem, (PR) ² = (PQ)² + (PR)² ⇒(QR)² = (13)² - (12) =169 – 144 = 25 QR = 5 cm tan p - cot r= QR/PQ - QR/PQ= 5/12 - 5/12=0 ...more

Using pythagoras theorem,

(PR) ² = (PQ)² + (PR)²

 

 

⇒(QR)² = (13)² - (12)

=169 – 144 = 25

 QR = 5 cm



tan p - cot r


= QR/PQ - QR/PQ

= 5/12 - 5/12

=0

 

Answers 3 Comments
Dislike Bookmark

Answered on 01/08/2019 Learn CBSE/Class 10/Mathematics/UNIT V: Trigonometry/Introduction to Trigonometry/NCERT Solutions/Exercise 8.1

In Δ ABC, right-angled at B Using Pythagoras theorem AC² = AB² +BC² AC² = 576 + 49 = 625 AC = √625 AC = ±25 Now (i) In a right angle triangle ABC where B=90° , Sin A = BC/AC =7/25 CosA =AB/AC =24/25 (ii) Sin C =AB/AC =24/25 Cos C =BC/AC =7/25 ...more

In Δ ABC, right-angled at B

Using Pythagoras theorem

AC² = AB² +BC²  

AC² = 576 + 49 = 625

AC = √625

AC = ±25

 

Now

(i) In a right angle triangle ABC where B=90° ,

Sin A = BC/AC

=7/25

CosA =AB/AC

=24/25

(ii) Sin C =AB/AC

=24/25

Cos C =BC/AC

=7/25

Answers 3 Comments
Dislike Bookmark

Shubhangi B. describes herself as Achieve excellence in Commerce with expert. She conducts classes in BBA Tuition, BCom Tuition and Class 10 Tuition. Shubhangi is located in Sector 61, Sahibzada Ajit Singh Nagar. Shubhangi takes Regular Classes- at her Home. She has 3 years of teaching experience . Shubhangi is pursuing Doctor of Philosophy (Ph.D.) from Panjab University . Shubhangi has completed Bachelor of Commerce (B.Com.) from Panjab University in 2017 and Master of Commerce (M.Com.) from Panjab University in 2019. She is well versed in English.

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