Question

In $△ABC,\angle B={90}^{\circ },BC=7$ and $AC-AB=1$, then $\cos C=.....$

• A. $\frac{3}{25}$
• B. $\frac{6}{25}$
• C. $\frac{7}{25}$
• D. $\frac{8}{25}$

Answer 1

Answer: (C)

$\frac{7}{25}$

In $△ABC$,

$AC-AB=1$

So, $AC=AB+1$

In $△ABC$,

$A{C}^2=A{B}^2+B{C}^2$ ...By Pythagoras theorem

$(AB+1{)}^2=A{B}^2+B{C}^2$

$\therefore 1+2AB+A{B}^2=A{B}^2+B{C}^2$
So, $1+2AB=(7{)}^2$

$\therefore 2AB=49-1=48$

$AB=\frac{48}{2}=24$ and $AC=1+AB=25$

$\cos C=\frac{BC}{AC}=\frac{7}{25}$

So, $\frac{7}{25}$ is correct.