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  • In triangle ABC if <img alt="a+c-\frac{3b}{2}=a \sin^2(\frac{C}{2})+c \sin^2 (\frac{A}{2})\\" src="https://learn.careers360.com/latex-image/?a&plus;c-%5Cfrac%7B3b%7D%7B2%7D%3Da%20%5Cs

In triangle ABC if a+c-\frac{3b}{2}=a \sin^2(\frac{C}{2})+c \sin^2 (\frac{A}{2})\\ then a,b,c are in ?

Option: 1

A.P.


Option: 2

G.P.


Option: 3

H.P.


Option: 4

None of these


Answers (1)

best_answer

 

 

Projection Formula -

Projection Formula

In the ΔABC, 

Projection of AB on BC is BD

Projection of AC on BC is DC

 

 

Now, 

BD = c cos B and DC = b cos C

and, BC = a = BD + DC

                    = c cos B + b cos C

a = c cos B + b cos C

In a similar way, other projection formula can be derived

  1. a = c cos B + b cos C

  2. b = c cos A + a cos C

  3. c = b cos A + a cos B

-

 

 

Diagram- 

a+c-\frac{3b}{2}=a \sin^2(\frac{C}{2})+c \sin^2 (\frac{A}{2})\\ a+c-\frac{3b}{2}=a[\frac{1- \cos C}{2} ]+c[\frac{1-\cos A}{2}] \\ c\cos A +a \cos C=3b-a-c\\ b=3b-a-c\\ \text{by above triangle } c\cos A +a \cos C=b\\ a+c=2b\\ a,b,c \ in\ A.P.

Posted by

shivangi.bhatnagar

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