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  • In triangle ABC if <img alt="\cos \frac{B}{2}=\sqrt{\frac{a+c}{2c}}\\" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Ccos%20%5Cfrac%7BB%7D%7B2%7D%3D%5Csqrt%7B%5Cfrac%7Ba&plus;c%7D%

In triangle ABC if \cos \frac{B}{2}=\sqrt{\frac{a+c}{2c}}\\, then angle C is ?

Option: 1

\frac{\pi}{6}


Option: 2

\frac{\pi}{4}


Option: 3

\frac{\pi}{3}


Option: 4

\frac{\pi}{2}


Answers (1)

best_answer

\\\cos \frac{B}{2}=\sqrt{\frac{a+c}{2c}}\\ \sqrt{\frac{s(s-b)}{ac}}=\sqrt{\frac{a+c}{2c}}\ \ \ \ where\ s=\frac{a+b+c}{2}\\ \frac{s(s-b)}{ac}=\frac{a+c}{2c}\\ 2s(s-b)=a(a+c)\\ (a+b+c)(a+c-b)=2a(a+c)\\ (a+c)^2-b^2=2a^2+2ac\\ a^2+b^2=c^2\\ \angle C=\frac{\pi}{2}

Posted by

Ritika Kankaria

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