Question

ABC is an isosceles right-angled triangle. Assuming AB = BC = x, find the value of each of the following trigonometric ratios : $tan{45}^{\circ }$

Answer 1

Given: $△ABC$, $AB=BC=x$, $\angle ABC={90}^{\circ }$
By Pythagoras Theorem,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=x^2+x^2$
$A{C}^2=2x^2$
$AC=\sqrt{2}x$
Since, $AB=BC$, then by Isosceles triangle property,
$\angle ACB=\angle CAB={45}^{\circ }$
Now, $tan{45}^{\circ }=tanC=\frac{P}{B}=\frac{AB}{BC}=\frac{x}{x}=1$