2. in , is a right angle. find the remaining sides and angles. round your answers to the nearest tenth. show your work. a = 3, c = 19

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°

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632