$BC=\sqrt{AC^2-AB^2}=\sqrt{26^2-15^2}=\sqrt{451}$ (Py-ta-go)
$\Rightarrow \widehat{A}=cos^{-1}.\left(\dfrac{15^2+26^2-\sqrt{451}^2}{2.15.26}\right) \approx 54^{o}45'56,1''$
$\widehat{C}=180^{o}-\left(54^{o}45'56,1''+90^{o}\right) \approx 35^{o}14'3,9''$
$\Rightarrow d_{AB}=\dfrac{2}{26+\sqrt{451}}.26.\sqrt{451}.cos\left(\dfrac{35^{o}14'3,9''}{2}\right) \approx 22,28178$
Áp dụng định lý hàm số cos:
$IA=\sqrt{26^2+22,28178^2-2.26.22,28178.cos\left(54^{o}45'56,1''\right)} \approx 22,45049$
Mình không chắc nhen !
