Answer: b=18.8; A=9.1°; B=80.9°; and C=90°
Solution:
Assuming that the hypothenuse is c=19 and one leg is a=3
Using the Phythagoras Theorem:
c^2=a^2+b^2
Replacing c=19 and a=3 in the equation above:
19^2=3^2+b^2
Squaring:
361=9+b^2
Solving for b. Isolating b^2: Subtracting 9 both sides of the equation:
361-9=9+b^2-9
352=b^2
b^2=352
Square root both sides of the equation:
sqrt(b^2)=sqrt(352)
b=18.76166303
Rounding to the nearest tenth:
b=18.8
Angles:
The opposite angle to the hypothenuse c must be a right angle (angle of 90°):
C=90°
Using the trigonometric function sine of the angle A:
sin A = (Opposite side to angle A) / hypothenuse
sin A=a/c
Replacing a=3 and c=19 in the equation above:
sin A=3/19
Solving for A:
A= sin^(-1) (3/19)
A=sin^(-1) (0.157894737)
A=9.084720297°
Rounding to the nearest tenth:
A=9.1°
Using that the acute angles in a right triangle are complementary (must add 90°):
A+B=90°
Replacing A=9.1° in the equation above:
9.1°+B=90°
Solving for B: Subtracting 9.1° both sides of the equation:
9.1°+B-9.1°=90°-9.1°
B=80.9°
