logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
0 votes

The angles $A,B$ and $C$ of a triangle ABC are in arithmetic progression.If $2b^2=3C^2$ determine the angle A

$(a)\;90^{\large\circ}\qquad(b)\;45^{\large\circ}\qquad(c)\;75^{\large\circ}\qquad(d)\;60^{\large\circ}$

Can you answer this question?
 
 

1 Answer

0 votes
Step 1:
Since $A,B,C$ are in A.P
$2B=A+C$
Also,$A+B+C=180^{\large\circ}$
$2B+B=18^{\large\circ}$
$3B=180^{\large\circ}$
$B=\large\frac{180^{\large\circ}}{3}$
$B=60^{\large\circ}$
Step 2:
Given :
$2b^2=3C^2$
$2k^2\sin^2B=3k^2\sin^2C$
$2\sin^260^{\large\circ}=3\sin^2C$
$2.\large\frac{3}{4}$$=3\sin^2C$
$\sin^2C=\large\frac{1}{2}$
$\sin C=\large\frac{1}{\sqrt 2}$
$C=\sin^{-1}\large\frac{1}{\sqrt 2}$
$C=45^{\large\circ}$
Step 3:
Hence $A=180^{\large\circ}-(B+C)$
On substituting B & C we get,
$A=180^{\large\circ}-(60^{\large\circ}+45^{\large\circ})$
$\;\;\;=75^{\large\circ}$
Hence (c) is the correct answer.
answered Oct 21, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...