Question

In ∆abc a=4, b=5, C=7, then since/2=?

Answer 1

Given :

For A Triangle, ABC

The value of a = 4

The value of b = 5

The value of c = 7

To Find :

The value of Sin$\dfrac{C}{2}$

Solution :

∵ For any Triangle, from Sin Law

Sin$\dfrac{C}{2}$  =  $\sqrt{\dfrac{\sqrt{(S-a)(S-b)} }{ab} }$                ............1

∵  S = $\dfrac{a+b+c}{2}$

        = $\dfrac{4+5+7}{2}$

        = $\dfrac{16}{2}$ = 8

Now, Put the value of S into eq 1

i.e  Sin$\dfrac{C}{2}$ =  $\sqrt{\dfrac{\sqrt{(S-a)(S-b)} }{ab} }$    

                = $\sqrt{\dfrac{(S-a)\times (S-b)}{a\times b}}$

                = $\sqrt{\dfrac{(8-4)\times (8-5)}{8\times 5}}$

                = $\sqrt{\dfrac{4\times 3}{8\times 5}}$

               = $\sqrt{\dfrac{ 3}{ 10}}$

Hence, The value of  Sin$\dfrac{C}{2}$  is $\sqrt{\dfrac{ 3}{ 10}}$  Answer