Question

In $\Delta ABC$, right angled at $A,{\cos }^{-1}\left(\frac{R}{r_2+r_3}\right)$ is

• A. ${30}^0$
• B. ${60}^0$
• C. ${90}^0$
• D. ${45}^0$

Answer 1

Answer: (B)

${60}^0$

$a^2=b^2+c^2$

$R=\frac{a}{2}$

$r_2=\frac{△}{(S-b)},r_3=\frac{△}{(S-c)}$

${\cos }^{-1}(\frac{a/2}{\frac{\frac{1}{2}bc}{(S-b)}+\frac{\frac{1}{2}bc}{(S-c)}})△=\frac{1}{2}bc$

$=[\frac{a(S-b)(S-c)}{bc(S-b+S-c)}]$

$=\frac{(\frac{a+c-b}{2})(\frac{a+b-c}{2})}{bc}$

$=\frac{a^2-b^2-c^2+2bc}{bc\times 4}$

$={\cos }^{-1}(\frac{1}{2})$

$={60}^o$.