In ΔABC, a = 8 and b = 8. If the area of ΔABC is 16, then m∠C may be equal to (1)30 (2)40 (3)50 (4)60?

It's 30 or 60 degrees because those and 45 degrees are the only ones we're ever expected to know.   Let's figure out which.

I'm not sure if this is trig.  The trig solution is easy,

area = (1/2) a b sin C

16 = (1/2) (8)(8) sin C

16/32 = sin C

sin C = 1/2

C = 30 degrees.

(1) 30

There's also C=150 degrees that works, but not among the choices.

It's harder if we don't have trig yet.  An isosceles triangle, like here with a=BC=b=AC, is bisected in two by the altitude to the unique base, let's call it h=CD where D is the midpoint of AB.

We're told the area of ABC is 16.  Euclid's Theorem for the area of a triangle is of course one half base times height.

area = 1/2 ch

16 = (1/2) ch

ch = 32

h=32/c = 2⁵/c

We're gonna need another equation.  The two resulting triangles are right triangles, ADC and BDC, both right angles at D (that's what an altitude is) and both with an unknown base AD=c/2 (where c=AB) and hypotenuse a=b=8.

By the Pythagorean Theorem,

(c/2)² + h² = a² = 8² = 64

c²/4 + h² = 8² = 2⁶

There's pretty hard going ahead; I'll stop here and assume this is a trig problem.

b.-4 units

all you   do is count how many times you move

70

step-by-step explanation: