Question

In right triangle ABC, $BC=7cm,AC-AB=1cm$ and $\angle B={90}^o$. The value of cos A + cos B+ cos C is

• A. $\frac{1}{7}$
• B. $\frac{32}{24}$
• C. $\frac{31}{25}$
• D. $\frac{25}{31}$

Answer 1

The correct option is C $\frac{31}{25}$

$A{C}^2=A{B}^2+49$
$(AC-AB)(AC+AB)=49$
$AC+AB=49$
$cosA+cosB+cosC=\frac{24}{25}+0+\frac{7}{25}=\frac{31}{25}$