Question

In the given figure; $BC=15$ cm and $\sin B=\frac{4}{5}$.
The measure of $AC$ in cm is

• A. $10$
• B. $15$
• C. $20$
• D. $19$

Answer 1

Answer: (C)

In $△ABC$, $BC=15$, $\sin B=\frac{4}{5}$

Now, $\sin B=\frac{P}{H}$
$\frac{4}{5}=\frac{AC}{AB}$
Then, let $AC=4x$ and $AB=5x$

Now, $A{B}^2=A{C}^2+B{C}^2$
$(5x{)}^2=(4x{)}^2+{15}^2$
$25x^2=16x^2+225$
$9x^2=225$
$x=5$

Hence, $AC=20$ and $AB=25$