solve each triangle. round each side length to the nearest tenth and angle measures to the nearest degree. a= 14, c= 20, B= 38 d
egrees.
2 answers:
A) the missing degree is 166
b) the missing degree is 142
c)the missing degree is 160
I do hope I helped you in a way! If I am incorrect I apologize.
Okay after scratching my head and some scrap paper :P I was really surprised to see that I don't remember ever running into this before...
Using the Law of Cosines we find the third side.
c=√(196+400-560cos38)≈12.4384 (we can round further for the final answer)
Now using the Law of Sines for the other angles.
sin38/12.4384=sinA/14=sinB/20
A=arcsin(14sin38/12.4384)≈43.8645°
BUT what I didn't realize before is that B was obtuse which means:
B=180-arcsin(20sin38/12.4384)≈98.135°
So solved for sides and angles (with rounding asked for)
(14, 44°), (12.4, 38°), and (20, 98°)
You might be interested in
Hopefully I drew what you described.
The 6.6 represents the total of N + 3.4
We subtract the part we know from the whole /sum/total we know to get the other part.
So 6.6 - 3.4 = N
3.2 = N
Answer:
C
Step-by-step explanation:
1/4= 0.25
Each curve completes one loop over the interval

. Find the intersections of the curves within this interval.

The region of interest has an area given by the double integral

equivalent to the single integral

which evaluates to

.
Answer:
x=9.4
x=9.43
Step-by-step explanation:
I used the Pythagorean theorem.
Have a nice day!
and can I maybe have brainliest? Thanks!
Answer:
it is a, b, and c, but not d