Question

Suppose an isosceles triangle ABC has A=pi/4 and b=c=4. What is the length of a^2?

Please explain. Thank you!

Answer 1

@jim_thompson5910 Does this look right?

c^2 = a^2 +b^2 -2ab*cos(C)
converted for A
a^2 = b^2 +c^2 -2bc*cos(c)
a^2 = 4^2 + 4^2 - 2*4*4*cos(pi/4)
a^2 = 32 - 32*sqrt(2)/2 = 32- 16*sqrt(2)
a^2= 32 - 16*sqrt(2)
4^2(2-sqrt(2) )= 32 - 16*sqrt(2)

so D) 4^2(2-√2) is the answer.

Answer 2

you have it correct

Answer 3

Thank you!

Answer 4

np