# Need help with a geometry question?

Mar 24, 2018

A=94.5°

B=92.5°

C=90.5°

D=82.5°

#### Explanation:

Let x equal the angle of $\textcolor{\mathmr{and} a n \ge}{B}$

Angle $\textcolor{red}{\angle} A$ = $x + 2$

Angle $\textcolor{g r e e n}{\angle} C$ = $x - 2$

Angle $\textcolor{b l u e}{\angle} D$ = $x - 10$

$\text{We know that the angle of any four-sided shape is equal to}$ color(purple)360°.

$\textcolor{red}{\angle A}$+$\textcolor{\mathmr{and} a n \ge}{\angle B}$+$\textcolor{g r e e n}{\angle C}$+$\textcolor{b l u e}{\angle D}$=360°

$\text{Substitute your values}$

$\left(x + 2\right)$ + $\left(x\right)$ + $\left(x - 2\right)$ + $\left(x - 10\right)$ $=$ 360°

$4 x - 10 = 360$

$4 x = 360 + 10$

$4 x = 370$

x=92.5°

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

Mar 24, 2018

#### 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 ${\textcolor{b l u e}{180}}^{\circ}$ 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

$\angle B A D = {\left(x + 2\right)}^{\circ}$

$\angle B C D = {\left(x - 2\right)}^{\circ}$

$\angle A D C = {\left(x - 10\right)}^{\circ}$

$\angle A B C =$ unavailable.

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

$\angle A B C + {\left(x - 10\right)}^{\circ} = {180}^{\circ}$. Equation 1

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

${\left(x + 2\right)}^{\circ} + {\left(x - 2\right)}^{\circ} = {180}^{\circ}$. Equation 2

Consider Equation 2 first.

${\left(x + 2\right)}^{\circ} + {\left(x - 2\right)}^{\circ} = {180}^{\circ}$

$\Rightarrow x + 2 + x - 2 = 180$

$\Rightarrow x + \cancel{2} + x - \cancel{2} = 180$

$\Rightarrow 2 x = 180$

Divide both side by 2

$\Rightarrow \frac{2 x}{2} = \frac{180}{2}$

$\Rightarrow \frac{\cancel{2} x}{\cancel{2}} = {\cancel{180}}^{\textcolor{red}{90}} / \cancel{2}$

Hence,

color(blue)(x= 90

So, when $x = 90$,

$\angle B A D = 90 + 2 = {92}^{\circ}$

$\angle B C D = 90 - 2 = {88}^{\circ}$

$\angle A D C = 90 - 10 = {80}^{\circ}$

We know that

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

$\Rightarrow \angle A B C + {80}^{\circ} = {180}^{\circ}$.

Subtract ${80}^{\circ}$ from both sides.

$\Rightarrow \angle A B C + {80}^{\circ} - {80}^{\circ} = {180}^{\circ} - {80}^{\circ}$.

$\Rightarrow \angle A B C + {\cancel{80}}^{\circ} - {\cancel{80}}^{\circ} = {180}^{\circ} - {80}^{\circ}$.

$\Rightarrow \angle A B C = {100}^{\circ}$.

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^@