Question

If $\alpha ,\beta ,\gamma$ are the lengths of the altitudes of $\Delta ABC$, then $\frac{\cos A}{\alpha }+\frac{\cos B}{\beta }+\frac{\cos C}{\gamma }=$

• A. $\Delta$
• B. $1/\Delta$
• C. R
• D. 1/R

Answer 1

The correct option is D 1/R

Let us assume that Given Triangle is an Equilateral Triangle

in which $\alpha =\beta =\gamma =asin{60}^o=a\frac{\sqrt{3}}{2}$

and $cosA=cosB=cosC=cos{60}^0=\frac{1}{2}$

As we know that Area of Equilateral Triangle $=\frac{\sqrt{3}}{4}{a}^2$

formula for Circum radius is $R=\frac{abc}{4△}=\frac{a^3}{4\ast \frac{\sqrt{3}}{4}\ast a^2}=\frac{a}{\sqrt{3}}$

Therefore, $\frac{cosA}{\alpha }+\frac{cosA}{\beta }+\frac{cosA}{\gamma }=3\ast \left(\frac{\frac{1}{2}}{a\frac{\sqrt{3}}{2}}\right)=\frac{\sqrt{3}}{a}=\frac{1}{R}$

Therefore Answer is $D$