Need help with a geometry question?

Let x equal the angle of #color(orange)B#

Angle #color(red)/_A# = #x+2#

Angle #color(green)/_C# = #x-2#

Angle #color(blue)/_D# = #x-10#

#"We know that the angle of any four-sided shape is equal to"# #color(purple)360°#.

#color(red)(/_A)#+#color(orange)(/_B)#+#color(green)(/_C)#+#color(blue)(/_D)#=360°

#"Substitute your values"#

#(x+2)# + #(x)# + #(x-2)# + #(x-10)# #=# #360°#

#4x-10=360#

#4x=360+10#

#4x=370#

#x=92.5°#

Substitute your x-value into A, C, and D.

Given:

Analyze the problem constructed using a geometry software available below:

Please note that the diagram is not drawn to scale.

We observe the following:

1. The quadrilateral ABCD is inscribed in a circle.

2. ABCD is a cyclic quadrilateral, since all the vertices of the quadrilateral touch the circumference of the circle.

Properties associated with angles in cyclic quadrilaterals:

The opposite angles of a cyclic quadrilateral add to #color(blue)180^@# or #color(red)(pi" radians"#.

We can use this useful property to solve our problem by chasing angles:

Hence,

#color(blue)(/_ ABC+/_ ADC = 180^@#

#color(blue)(/_ BAD+/_ BCD = 180^@#

Given that

#/_BAD=(x+2)^@#

#/_BCD=(x-2)^@#

#/_ADC=(x-10)^@#

#/_ABC= # unavailable.

As, #color(blue)(/_ ABC+/_ ADC = 180^@#,

#/_ ABC+(x - 10)^@ = 180^@#. Equation 1

As, #color(blue)(/_ BAD+/_ BCD = 180^@#,

#(x+2)^@ + (x-2)^@=180^@#. Equation 2

Consider Equation 2 first.

#(x+2)^@ + (x-2)^@=180^@#

#rArr x+2+x-2=180#

#rArr x+cancel 2+x-cancel 2=180#

#rArr 2x=180#

Divide both side by 2

#rArr (2x)/2=180/2#

#rArr (cancel2x)/cancel 2=cancel 180^color(red)(90)/cancel 2#

Hence,

#color(blue)(x= 90#

So, when #x=90#,

#/_ BAD = 90+2 =92^@#

#/_ BCD = 90-2 =88^@#

#/_ ADC = 90-10 =80^@#

We know that

#color(blue)(/_ ABC+/_ ADC = 180^@#.

#rArr /_ ABC+80^@= 180^@#.

Subtract #80^@# from both sides.

#rArr /_ ABC+80^@-80^@= 180^@-80^@#.

#rArr /_ ABC+cancel 80^@-cancel 80^@= 180^@-80^@#.

#rArr /_ ABC= 100^@#.

Now, we are in a position to write all our angles as follows:

#color(green)(/_ BAD = 92^@;/_ BCD = 88^@; /_ ADC = 80^@; /_ ABC= 100^@#.

Next, let us verify all the four angles add to #color(red)(360^@#

#/_ BAD + /_ BCD +/_ ADC+ /_ ABC = 92^@+88^@+80^@+100^@ = color(red)(360^@#