Answer:
![m\angle A=\tan^{-1}(\frac{m}{n})](https://tex.z-dn.net/?f=m%5Cangle%20A%3D%5Ctan%5E%7B-1%7D%28%5Cfrac%7Bm%7D%7Bn%7D%29)
![AB=\sqrt{m^2+n^2}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7Bm%5E2%2Bn%5E2%7D)
Step-by-step explanation:
Given:
In Δ ABC
m∠C= 90°
![BC=m](https://tex.z-dn.net/?f=BC%3Dm)
![AC=n](https://tex.z-dn.net/?f=AC%3Dn)
For the triangle ABC we can apply trigonometric ratio to find m∠A.
We have:
![\tan\angle A=\frac{BC}{AC}](https://tex.z-dn.net/?f=%5Ctan%5Cangle%20A%3D%5Cfrac%7BBC%7D%7BAC%7D)
Substituting values of side BC and AC
![\tan \angle A=\frac{m}{n}](https://tex.z-dn.net/?f=%5Ctan%20%5Cangle%20A%3D%5Cfrac%7Bm%7D%7Bn%7D)
Taking inverse tan to get m∠A.
∴ ![m\angle A=\tan^{-1}(\frac{m}{n})](https://tex.z-dn.net/?f=m%5Cangle%20A%3D%5Ctan%5E%7B-1%7D%28%5Cfrac%7Bm%7D%7Bn%7D%29)
AB can be found out using Pythagorean theorem:
AB being the hypotenuse can be written as
![AB^2=BC^2+AC^2](https://tex.z-dn.net/?f=AB%5E2%3DBC%5E2%2BAC%5E2)
Substituting values of side BC and AC
![AB^2=m^2+n^2](https://tex.z-dn.net/?f=AB%5E2%3Dm%5E2%2Bn%5E2)
Taking square roots both sides:
![\sqrt{AB^2}=\sqrt{m^2+n^2}](https://tex.z-dn.net/?f=%5Csqrt%7BAB%5E2%7D%3D%5Csqrt%7Bm%5E2%2Bn%5E2%7D)
∴ ![AB=\sqrt{m^2+n^2}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7Bm%5E2%2Bn%5E2%7D)
Answer:
z=9
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
Answer:
44 in^2
Step-by-step explanation:
break it into 2 smaller rectangles and solve those.