Question

In $\Delta$ ABC, AD is perpendicular to BC, then $\frac{AC}{AB}$ is equal to

• A. $\frac{sinC}{sinB}$
• B. $\frac{cosB}{cosC}$
• C. $\frac{sinB}{sinC}$
• D. $\frac{tanB}{tanC}$

Answer 1

The correct option is C $\frac{sinB}{sinC}$

$GivenAD\perp BC$

$\therefore In△ABD$

$\frac{AD}{AB}=sinB---\left(1\right)$

$andin△ADC$

$\frac{AD}{AC}=sinC---\left(2\right)$

$Dividing\left(1\right)by\left(2\right)weget$

$\frac{AC}{AB}=\frac{sinB}{sinC}$