Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Muskan Malik 6 years, 2 months ago
- 2 answers
Posted by Naruto Uzumaki??? 6 years, 2 months ago
- 2 answers
Naruto Uzumaki??? 6 years, 2 months ago
Posted by Krishika Chaturvedi 6 years, 2 months ago
- 0 answers
Posted by Akhilesh Singh Kushwah 6 years, 2 months ago
- 2 answers
Rohit Kumar 6 years, 2 months ago
Posted by Himanshu Shahi 6 years, 2 months ago
- 5 answers
Posted by Ramdas Murmu 6 years, 2 months ago
- 0 answers
Posted by Sanjay Bhatt 6 years, 2 months ago
- 1 answers
Ram Kushwah 6 years, 2 months ago
15tanA=14
tanA =14/15
tan2A=2tan2A/1-tan2A
=2*(14/15)/(1-225/196)
tan2A=28/15/29/225=420/29
=======================================
Now 2sinAcosA/cos2A-sin2A
=sin2A/cosA=tan2A
=420/29
Posted by Rajeev Kumar Singh 6 years, 2 months ago
- 1 answers
Ram Kushwah 6 years, 2 months ago
We know that area of an equilateral triangle with side a
={tex}\sqrt3{/tex} a2/4
now let side of square =x
then area of equilateral triangle on one side
A1={tex}\sqrt3{/tex} x2/4--------------------------(1)
If length of diagonal is d then
d2=x2+x2=2x2
area of equilateral traingle at diagonal
A2=={tex}\sqrt3{/tex} *2x2/4
=2*({tex}\sqrt3{/tex} a2/4)
=2A1
hence proved
Posted by Shaad Saifi 6 years, 2 months ago
- 4 answers
Posted by Sahil Prasad 6 years, 2 months ago
- 1 answers
Aadya Singh ? 6 years, 2 months ago
Posted by Suhani Semwal 6 years, 2 months ago
- 4 answers
Rishi Sharma 6 years, 2 months ago
Posted by Laranya Dwivedi 6 years, 2 months ago
- 0 answers
Posted by Akshit Panchal 6 years, 2 months ago
- 0 answers
Posted by Akshit Panchal 6 years, 2 months ago
- 5 answers
Posted by Monika Sharma 6 years, 2 months ago
- 1 answers
Qwerty G 6 years, 2 months ago
Posted by Vedang Nain 6 years, 2 months ago
- 3 answers
Qwerty G 6 years, 2 months ago
Rufra Pratap Singh Rathore 6 years, 2 months ago
Vedang Nain 6 years, 2 months ago
Posted by Harsh Sethi 6 years, 2 months ago
- 0 answers
Posted by Prasanna Prasanna 6 years, 2 months ago
- 2 answers
Rohit Kumar 6 years, 2 months ago
Posted by Rupesh Joshi 6 years, 2 months ago
- 0 answers
Posted by Shubhranshu Jena 6 years, 2 months ago
- 0 answers
Posted by Aryan Dixit 6 years, 2 months ago
- 3 answers
Bikram Bal 6 years, 2 months ago
Yogita Ingle 6 years, 2 months ago
Ratios of Complementary angles ,
Two angles are said to be Complementary , if their sum is equal
to 90°.
As you know that ,
Cot ( 90 - x ) = tan x
According to the problem given,
Tan 2A = cot ( A - 18 )
Cot ( 90 - 2A ) = cot ( A - 18 )
Remove the cot both sides
we get ,
90 - 2A = A - 18
-2A - A = - 18 - 90
- 3A = - 108
A = ( - 108 ) / ( - 3 )
A = 36°
Posted by Safaqnaz Khan 6 years, 2 months ago
- 7 answers

myCBSEguide
Trusted by 1 Crore+ Students

Test Generator
Create papers online. It's FREE.

CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
myCBSEguide
Akshit ???? 6 years, 2 months ago
0Thank You