Question

In triangle $ABC$, the medians intersect at point $G$. Suppose $\angle ABC={90}^o$ and $\angle BAC+\angle BGC={180}^o$. Find the value of $\cos \angle ACB$.

• A. $\frac{\sqrt{2}}{\sqrt{3}}$
• B. $\frac{\sqrt{2}}{\sqrt{5}}$
• C. $\frac{2}{\sqrt{5}}$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

Let $\angle BAC=Q$

$\Rightarrow \angle BAC+\angle BGC=180$

$\Rightarrow Q+2Q=180$

$\Rightarrow Q=60°=\angle BAC$

$\Rightarrow \angle ACB=90°-60°=30°$

$\Rightarrow \cos 30°=\frac{\sqrt{3}}{2}$

Hence, the answer is $\frac{\sqrt{3}}{2}.$