Law of cosines:
a^2=b^2+c^2 -2acCosB
=3+10-2sqrt(3*10)*3/sqrt10
=13-6sqrt3
finish it for solving a.
check my math.
I have no idea how to do this!!!
Let ΔABC be a triangle such that b=sqrt(3), c=sqrt(10), and cos(B)=3/sqrt(10). Find all possible values for side length a.
Please show work
3 answers
sinB = 1/√10
b/sinB = c/sinC, so
√3/(1/√10) = √10/sinC
sinC = 1/√3
A = 180 - (B+C)
sinA = (180 - (B+C)) = sin(B+C)
= sinBcosC+cosBsinC
= (1/√10)(√2/√3)+(3/√10)(1/√3)
= (2+3√2)/(2√15)
Now get
a/sinA = b/sinB
or
a^2 = b^2+c^2 - 2bc*cosA
b/sinB = c/sinC, so
√3/(1/√10) = √10/sinC
sinC = 1/√3
A = 180 - (B+C)
sinA = (180 - (B+C)) = sin(B+C)
= sinBcosC+cosBsinC
= (1/√10)(√2/√3)+(3/√10)(1/√3)
= (2+3√2)/(2√15)
Now get
a/sinA = b/sinB
or
a^2 = b^2+c^2 - 2bc*cosA
a^2=b^2+c^2-2bc*cosA
and we don't know A yet
and we don't know A yet
1 answer