SOLUTION: The area of a triangle is 40 cm squared. Find the length of the side included between the angles A=30 degrees and B=50 degrees.

Algebra ->  Triangles -> SOLUTION: The area of a triangle is 40 cm squared. Find the length of the side included between the angles A=30 degrees and B=50 degrees.      Log On


   

Question 934849: The area of a triangle is 40 cm squared. Find the length of the side included between the angles A=30 degrees and B=50 degrees.
Found 2 solutions by MathLover1, KMST:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Let
b = base of the triangle
h = altitude of the triangle
Area of a triangle A=+%281%2F2%29bh+=40
Therefore, b+=+2%2A40%2Fh or
b+=+80%2Fh ........eq. 1
Angle C+=+180+-+30+-+50+=+100
Using the Law of Sines,
sin+%28100%29%2F%28AB%29+=+sin+%2850%29%2Fb
where
AB = length of the side included between the angles A+=+30 and B+=50
Solving for "b",
b+=+%28AB%29%28sin+%2850%29%2Fsin+%28100%29%29 ............eq. 2
Since eq.1 = eq.2, then
80%2Fh+=+%28AB%29%28sin+%2850%29%2Fsin+%28100%29%29
Solving for AB,
OR h+=+AB%28sin+%2830%29%29+=+AB%2F2
since h+=+AB%2F2, substitute this in eq.3,
AB+=+80%28sin+%28100%29%29%2F%28%28AB%2F2%29%2Asin%2850%29%29
AB+=+2%2A80%28sin+%28100%29%29%2F%28AB%2Asin%2850%29%29


Simplifying the above,
%28AB%29%5E2+=+160%28sin+%28100%29%2Fsin+%2850%29%29
%28AB%29%5E2+=+160%280.9848%2F0.7660%29
%28AB%29%5E2+=+205.70
AB+=+14.34cm


Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let a, b . and c be the sides' lengths, in cm.
area%5BABC%5D=%281%2F2%29%2Ac%2Ab%2Asin%28A%29-->area%5BABC%5D=%281%2F2%29%2Ac%2Ab%2Asin%2830%5Eo%29-->40=%281%2F2%29c%2Ab%2A%281%2F2%29-->40=c%2Ab%2F4
C=180%5Eo-30%5Eo-50%5Eo--->C=100%5Eo
Law of sines says that
b%2Fsin%28B%29=c%2Fsin%28C%29 ,
so b%2Fsin%2850%5Eo%29=c%2Fsin%28100%5Eo%29--->b=c%2Asin%2850%5Eo%29%2Fsin%28100%5Eo%29
system%2840=c%2Ab%2F4%2Cb=c%2Asin%2850%5Eo%29%2Fsin%28100%5Eo%29%29 ---> 40=c%2Ac%2Asin%2850%5Eo%29%2F4sin%28100%5Eo%29--->40=c%5E2%2Asin%2850%5Eo%29%2F4sin%28100%5Eo%29--->c%5E2=4%2A40sin%28100%5Eo%29%2Fsin%2850%5Eo%29--->c%5E2=160sin%28100%5Eo%29%2Fsin%2850%5Eo%29
So, c=sqrt%28160sin%28100%5Eo%29%2Fsin%2850%5Eo%29%29--->c=14.342 (rounded)