Question

If $△$ ABC is an acute angled triangle and BD is perpendicular to AC, then the value of $\frac{\sin A}{\sin C}$

• A. $\frac{AB}{BC}$
• B. $\frac{BC}{AB}$
• C. $\frac{BC}{CD}$
• D. $\frac{CD}{BC}$

Answer 1

The correct option is B $\frac{BC}{AB}$

Given In $△ABC$ $BD\perp AB$

From $△BAD,$

$\sin A=\frac{BD}{AB}$

From $△BDC,$

$\sin C=\frac{BD}{BC}$

$\frac{\sin A}{\sin C}=\frac{\frac{BD}{AB}}{\frac{BD}{BC}}=\frac{BC}{AB}$