Question

why are dry erase boards so great?
i. find the desired angle measures given one angle.
if $\angle 2$ is $32^{\circ}$...
(a) $\angle 6 =$___$^{\circ}$
(e) $\angle 8 =$___$^{\circ}$
if $\angle 3$ is $27^{\circ}$...
(h) $\angle 6 =$___$^{\circ}$
(y) $\angle 5 =$___$^{\circ}$
if $\angle 4$ is $114^{\circ}$...
(r) $\angle 5 =$___$^{\circ}$
(e) $\angle 6 =$___$^{\circ}$
if $\angle 7$ is $41^{\circ}$...
(a) $\angle 1 =$___$^{\circ}$
(b) $\angle 3 =$___$^{\circ}$
if $\angle 5$ is $110^{\circ}$...
(e) $\angle 3 =$___$^{\circ}$
(t) $\angle 1 =$___$^{\circ}$
if $\angle 6$ is $64^{\circ}$...
(r) $\angle 3 =$___$^{\circ}$
(e) $\angle 4 =$___$^{\circ}$
ii. find the value of $x$
(l) $x =$___
(m) $x =$___
(r) $x =$___
(k) $x =$___

Explanation:

Step1: Find $\angle6$ when $\angle2 = 32^{\circ}$

Since $\angle2$ and $\angle6$ are corresponding angles (parallel lines cut by a transversal), $\angle6=\angle2$.
$$\angle6 = 32^{\circ}$$

Step2: Find $\angle8$ when $\angle2 = 32^{\circ}$

$\angle2$ and $\angle4$ are vertical angles ($\angle4 = 32^{\circ}$), and $\angle4$ and $\angle8$ are corresponding angles. So $\angle8=\angle4$.
$$\angle8 = 32^{\circ}$$

Step3: Find $\angle6$ when $\angle3 = 27^{\circ}$

$\angle3$ and $\angle5$ are supplementary ($\angle3+\angle5 = 180^{\circ}$), so $\angle5=180 - 27=153^{\circ}$. $\angle5$ and $\angle6$ are vertical angles, so $\angle6=\angle5$.
$$\angle6 = 153^{\circ}$$

Step4: Find $\angle5$ when $\angle3 = 27^{\circ}$

$\angle3$ and $\angle5$ are supplementary.
$$\angle5=180 - 27 = 153^{\circ}$$

Step5: Find $\angle5$ when $\angle4 = 114^{\circ}$

$\angle4$ and $\angle5$ are supplementary.
$$\angle5=180 - 114=66^{\circ}$$

Step6: Find $\angle6$ when $\angle4 = 114^{\circ}$

$\angle4$ and $\angle6$ are alternate - interior angles (parallel lines cut by a transversal).
$$\angle6 = 114^{\circ}$$

Step7: Find $\angle1$ when $\angle7 = 41^{\circ}$

$\angle7$ and $\angle5$ are vertical angles ($\angle5 = 41^{\circ}$), $\angle5$ and $\angle1$ are corresponding angles.
$$\angle1 = 41^{\circ}$$

Step8: Find $\angle3$ when $\angle7 = 41^{\circ}$

$\angle7$ and $\angle5$ are vertical angles ($\angle5 = 41^{\circ}$), $\angle5$ and $\angle3$ are supplementary.
$$\angle3=180 - 41 = 139^{\circ}$$

Step9: Find $\angle3$ when $\angle5 = 110^{\circ}$

$\angle5$ and $\angle3$ are supplementary.
$$\angle3=180 - 110 = 70^{\circ}$$

Step10: Find $\angle1$ when $\angle5 = 110^{\circ}$

$\angle5$ and $\angle1$ are corresponding angles.
$$\angle1 = 110^{\circ}$$

Step11: Find $\angle3$ when $\angle6 = 64^{\circ}$

$\angle6$ and $\angle4$ are vertical angles ($\angle4 = 64^{\circ}$), $\angle4$ and $\angle3$ are supplementary.
$$\angle3=180 - 64 = 116^{\circ}$$

Step12: Find $\angle4$ when $\angle6 = 64^{\circ}$

$\angle6$ and $\angle4$ are vertical angles.
$$\angle4 = 64^{\circ}$$

Step13: Find $x$ for $\angle L$

Since $(2x + 5)$ and $(4x+3)$ are alternate - interior angles (parallel lines cut by a transversal), $2x + 5=4x+3$.
$$2x+5 = 4x + 3$$
$$5 - 3=4x - 2x$$
$$2x=2$$
$$x = 1$$

Step14: Find $x$ for $\angle M$

Since $(5x - 7)$ and $(3x - 5)$ are supplementary (parallel lines cut by a transversal), $(5x - 7)+(3x - 5)=180$.
$$5x-7 + 3x-5=180$$
$$8x-12 = 180$$
$$8x=192$$
$$x = 24$$

Step15: Find $x$ for $\angle R$

Since $(6x - 20)$ and $(2x + 40)$ are alternate - interior angles (parallel lines cut by a transversal), $6x-20=2x + 40$.
$$6x-20=2x + 40$$
$$6x-2x=40 + 20$$
$$4x=60$$
$$x = 15$$

Step16: Find $x$ for $\angle K$

Since $(9x + 2)$ and $(5x + 10)$ are vertical angles, $9x+2=5x + 10$.
$$9x+2=5x + 10$$
$$9x-5x=10 - 2$$
$$4x=8$$
$$x = 2$$

Answer:

A. $32$; E. $32$; H. $153$; Y. $153$; R. $66$; E. $114$; A. $41$; B. $139$; E. $70$; T. $110$; R. $116$; E. $64$; L. $1$; M. $24$; R. $15$; K. $2$