SOLUTION: solve triangle ABC using the given info... a=4, b=6, c=9. please help me!!! and if you could explain that would be great, im so lost!

Question 32490: solve triangle ABC using the given info...
a=4, b=6, c=9. please help me!!! and if you could explain that would be great, im so lost!
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
USE FIRST a^2=b^2+c^2-2bcCOS(A)..
COS(A)=(b^2+c^2-a^2)/(2bc)
TO FIND ANGLE A AND THEN FOLLOW THE EXAMPLE GIVEN BELOW

SEE THE FOLLOWING EXAMPLE WHICH IS SAME AS YOUR QUESTION AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
please help me with this long question. i PROMISE it really is only one question in my book, but its super long. please believe me that its only one question...
Given triangle ABC with C=48 degrees, a=12 and b=8:
a. Solve for c, rounding your answer first to the nearest tenth and then to the nearest hundredth.
b. Use the law of sines with your first answer for c to find A
c. use the law of sines with your second answer for c to find A
d. Use the law of cosines to find angle A using your first answer for c
e. use the law of cosines to find angle A using your second answer for c
f. Find angle B using the two measures found in part a.
g. is angle A acute or abtuse and why?
im so sorry this is so long, its seriously how it is in the book... its a C excersise and they're usually the hard ones, and i dont get it at all!
1 solutions
Answer 19063 by venugopalramana(1346) About Me on 2006-04-05 12:11:24 (Show Source):
Given triangle ABC with C=48 degrees, a=12 and b=8:
a. Solve for c, rounding your answer first to the nearest tenth and then to the nearest hundredth.
C^2=a^2+b^2-2ab*COS(C),SUBSTITUTING THE ABOVE VALUES WE GET
c=8.9.....
c=8.92
b. Use the law of sines with your first answer for c to find A
a/SINA(A) = c/SIN(C)......
WITH c=8.9......NO ANSWER...AS VALUE EXCEEDS RANGE OF SIN(A)
c. use the law of sines with your second answer for c to find A
WITH c=8.92......A=88.71 DEG.
d. Use the law of cosines to find angle A using your first answer for c
COS(A)=(b^2+c^2-a^2)/2bc......A=90.3 DEG.WITH c=8.9
e. use the law of cosines to find angle A using your second answer for c
COS(A)=(b^2+c^2-a^2)/2bc......A=90.3 DEG.WITH c=8.92
f. Find angle B using the two measures found in part a.
b/SIN(B)=c/SIN(C)......B=41.9..WITH c=8.9...AND B=41.8...WITH c=8.92
g. is angle A acute or abtuse and why?
FOR ACTUAL VALUE OF c WE GET A=89.8...SO IT IS ACUTE.
im so sorry this is so long, its seriously how it is in the book... its a C excersise and they're usually the hard ones, and i dont get it at all!
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