Given a right angle triangle ABC
As shown in the figure:
The hypotenuse = BC = 19
One of the legs = AB = 10
We will use the Pythagorean theorem to find the other leg:
[tex]\begin{gathered} AB^2+AC^2=BC^2 \\ 10^2+AC^2=19^2 \\ AC^2=19^2-10^2 \\ AC^2=361-100=261 \\ AC=\sqrt[]{261}=16.16 \end{gathered}[/tex]
Now, we will find the measure of the angle B:
[tex]\begin{gathered} \cos B=\frac{adjacent}{hypotenuse}=\frac{10}{19} \\ B=\cos ^{-1}\frac{10}{19}\approx58.24 \end{gathered}[/tex]
The sum of the angles of the triangle = 180
so, the measure of angle C will be =
[tex]\begin{gathered} 180-(\angle A+\angle B) \\ =180-(90+58.24) \\ =180-148.24 \\ =31.76 \end{gathered}[/tex]
So, as a conclusion to the answer:
[tex]\begin{gathered} AC=16.16 \\ \angle B=58.24 \\ \angle C=31.76 \end{gathered}[/tex]
