Question

ABC is a triangle such that $m_a$ >$\frac{a}{2}$ where $m_a=$length of the median through A and $a=$ length of BC. Then

• A. Such a triangle does not exist
• B. $\angle A$ is acute.
• C. $\angle A$ is obtuse.
• D. $\angle A$ is right angle.
• E. none of these.

Answer 1

Answer: (B)

$\angle A$ is acute.

Length of median $m_a=\frac{\sqrt{2c^2+2b^2-a^2}}{2}$

$m_a>\frac{a}{2}\Rightarrow \frac{\sqrt{2c^2+2b^2-a^2}}{2}$

$\Rightarrow \sqrt{2c^2+2b^2-a^2}>a$

$\Rightarrow c^2+b^2>a^2$

$therefore,CosA>0(CosA=\frac{b^2+c^2-a^2}{2bc})$

$\Rightarrow$ $\angle A$ is acute angle.