I reallY need help with these problems. if somebody could give me the answers that would be great

∠AMC  = 75

Step-by-step explanation:

ΔABC,

∠C = 90°  ; ∠B = 30°

∠A + ∠B +∠C  = 180   {Angle sum property of triangle}

∠A + 30 + 90 = 180

       ∠A + 120 = 180

                 ∠A = 180 - 120

                 ∠A  = 60°

In ΔACM,

∠ACM = 90/2  =  45°         { CM is angle bisector}

∠ACM + ∠AMC +∠A =  180   {angle sum property}

   45 +  ∠AMC + 60 = 180

            ∠AMC + 105 = 180

                     ∠AMC = 180 - 105

∠AMC  = 75

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180