Solve right triangles can someone please answer if you do show your work

(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

18 units²

step-by-step explanation:

first let's find altitude bd using the distance formula

d = √((y₂ - y₁)² + (x₂ - x₁)²)

bd = √((0 - -3)² + (0 - 3)²)

bd = √18

given the placement of points a and c, we can denote that point d bisects segment ac.

we will call ac the base.

using the distance formula

ac = √((0 - 6)² + (-6 - 0)²)

ac = √72

knowing our base and our height, we can plug these into our triangle area formula

area = (b * h)/2

area = (√18 * √72) / 2

area = (√1296) / 2

area = 36 / 2

area = 18 units²

the answer is a and d