Answer:
[tex]m\angle A=\tan^{-1}(\frac{m}{n})[/tex]
[tex]AB=\sqrt{m^2+n^2}[/tex]
Step-by-step explanation:
Given:
In Δ ABC
m∠C= 90°
[tex]BC=m[/tex]
[tex]AC=n[/tex]
For the triangle ABC we can apply trigonometric ratio to find m∠A.
We have:
[tex]\tan\angle A=\frac{BC}{AC}[/tex]
Substituting values of side BC and AC
[tex]\tan \angle A=\frac{m}{n}[/tex]
Taking inverse tan to get m∠A.
∴ [tex]m\angle A=\tan^{-1}(\frac{m}{n})[/tex]
AB can be found out using Pythagorean theorem:
AB being the hypotenuse can be written as
[tex]AB^2=BC^2+AC^2[/tex]
Substituting values of side BC and AC
[tex]AB^2=m^2+n^2[/tex]
Taking square roots both sides:
[tex]\sqrt{AB^2}=\sqrt{m^2+n^2}[/tex]
∴ [tex]AB=\sqrt{m^2+n^2}[/tex]
