Question
Mathematics Question on Triangles
In a triangle ABC, BC=7, AC=8, AB=α∈N and cosA=32. If 49cos(3C)+42=nm, where gcd(m,n)=1, then m+n is equal to ________.
Answer
Using the cosine rule for cosA:
cosA=2bcb2+c2−a2
Substitute b=8, c=7, and cosA=32:
32=2×8×782+72−a2
⟹a2=9
⟹a=3
Now, calculate cosC using the cosine rule:
cosC=2×7×872+82−92=72
Then, for cos(3C), we use the triple angle formula:
49cos(3C)+42=49(4cos3C−3cosC)+42
Substituting cosC=72:
=49(4(72)3−3⋅72)+42
=49(34332−76)+42
=732+42
Thus, m=32 and n=7, so:
m+n=32+7=39