Quadrilateral ABCD is inscribed in a circle with angle measures m
*PLEASE ANSWER IMMEDIATELY*

B and D

Step-by-step explanation:

First method :

All angles of a quadrilateral sums to 360 degrees

A + B + C + D = 360

=> 5x^2 + 36x - 336 = 0

=> x = 6 or x = - 56/5 [It gives negative value of angles]

So, A = 58, B =  109, C = 122, D = 71

Second method :

Sum of opposite angles of a quadrilateral inscribed in a circle is 180 degrees

A + C = 180

=> x = 6

So, A = 58, B =  109, C = 122, D = 71

[If there's any mistake, please inform me]

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

56

step-by-step explanation: