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 $c^{2}-3 c-7=0$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.