Find the measure of AngleJ, the smallest angle in a triangle with sides measuring 11, 13, and 19. Round to the nearest whole degree. 30° 34° 42° 47°

  34°

Step-by-step explanation:

The law of cosines is good for finding angles when only sides are known. We'll use the conventional sides a, b, c, and angles A, B, C. Yes, we know the problem statement calls the smallest angle "J". We trust you can make the translation.

  a² = b² +c² -2bc·cos(A) . . . . . for sides a, b, c and angle A

Solving for the angle, we get ...

  A = arccos((b² +c² -a²)/(2bc))

Filling in the numbers with "a" being the shortest side, we have ...

  A = arccos((13² +19² -11²)/(2·13·19)) = arccos(409/494)

  A ≈ 34.113°

The smallest angle, ∠J, is about 34°.

b 1: 45 am

step-by-step explanation:

its that because if it was in the pm it would be 1345

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2