# In a ∆ABC, if a = 4, b = 3,

Question:

In a $\triangle A B C$, if $a=4, b=3, A=\frac{\pi}{3}$. Then side $C$ is given by

Solution:

In ∆ABC

if $a=4, b=3, A=\frac{\pi}{3}$

Since $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$

i. e $\cos \frac{\pi}{3}=\frac{9+c^{2}-16}{2 \times 3 \times c}$

i. e $\frac{1}{2}=\frac{c^{2}-7}{6 c}$

i. e $3 c=c^{2}-7$

i. e $c^{2}-3 c-7=0$