Question 32493: 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!
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 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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