Question

In triangle $ABC$; $\angle ABC={90}^o,\angle CAB=x^o,\tan x^o=\frac{3}{4}$ and $BC=15cm$. Find the measure of $AB$ and $AC$.

Answer 1

R.E.F.Image.

$Tanx=3/4$

$BC/AB=3/4$

$15/AB=3/4$

$3AB=60$

$AB=20cm$

In right $\Delta ABC$

$A{C}^2=A{B}^2+B{c}^2$ [by Pythagoras]

$={20}^2+{15}^2$

$=400+225=625$

$AC=\sqrt{625}=25cm$

$\therefore AB=20cm;AC=25cm.$