Question

Need help with a geometry question?

Answer 1

`A=94.5°`

`B=92.5°`

`C=90.5°`

`D=82.5°`

Explanation:

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.

Answer 2

Please read the explanation.

Explanation:

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^@`