Question

In triangle ABC; $\angle ABC={90}^{\circ },\angle CAB=x^{\circ },tan{x}^{\circ }=\frac{3}{4}$ and BC = 15 cm. the measures of AB in cm

Answer 1

Given, $\angle ABC={90}^{\circ }$, $BC=15$
$\tan \angle CAB=\frac{3}{4}$
Now, $\tan \angle CAB=\frac{P}{B}$
$\frac{3}{4}=\frac{BC}{AB}$
$AB=\frac{4\times 15}{3}$
$AB=20$
Now, Using Pythagoras Theorem,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2={20}^2+{15}^2$
$A{C}^2=625$
$AC=25$