1. HUH?

first off see attachment

this is not for my homework, i am a college math tutor

my mother is a math teacher, she teaches at ECSU and MMS

but, i have no clue how the book got P Q or R

i got S and T rather easily, and made an assumption ( i know forbidden in geometery...) that lines QP and TT (the line other than TS that extends from point T) were parallel

they are not, i later found out that the bent hash marks made on the lines denote parallel

what are your answers to P Q and R, my mom even asked a few of her teacher buddies and they came up with much the same thing i got (close but still off)

i will give the book answer and my answer in a later post

2. Originally Posted by buggerzapper
first off see attachment

this is not for my homework, i am a college math tutor

my mother is a math teacher, she teaches at ECSU and MMS

but, i have no clue how the book got P Q or R

i got S and T rather easily, and made an assumption ( i know forbidden in geometery...) that lines QP and TT (the line other than TS that extends from point T) were parallel

they are not, i later found out that the bent hash marks made on the lines denote parallel

what are your answers to P Q and R, my mom even asked a few of her teacher buddies and they came up with much the same thing i got (close but still off)

i will give the book answer and my answer in a later post
Without some dimesion to constrain the sides, you can only give a minimum & maximum values for the angles.

Or the "bent hash" marked lines are equal in distance as well as being parallel.

Something is missing.

There must be some additional information that limits the angles.

.

3. that's what we finally said in the end "lack of data."

it's odd though because this is in a book for practicing for the local state standardized test....

P=170
Q=118
R=135
S=90
T=135

which would require angle R and the other angle next to P to be marked as the same at T
and line QP to be marked as parallel to the line extending away from angle T and not going to S

the book said:

P=172
Q=116
R=137
S=90
T=135

but where they got those particular angles for P Q and R are just not doable without more infomation

4. Hello, buggerzapper!

Code:
     A                     B
o - - - - - - - - - o
| 90°            p *
|                 * 55°          F
|              C o- - - - - - - o
| 107°          *           90° |
E o              *                |
|   *         *                 |
|       *  q *                t |
|         D o r                 o G
|           |                 *
|           |               *
| 90°     s | 90°      t  *
o - - - - - o - - - - - o
J            I            H

We are given: . $AJ \parallel DI \parallel FG$

and: . $\angle A \,=\,\angle F \,=\angle J \,=\,\angle s \,=\,\angle DIH \,=\,90^o \quad\Rightarrow\quad \boxed{\angle s \,=\,90^o}$

Assuming that $AB \parallel CF,\:\angle p \,=\,\angle BCF \,=\,55^o \quad\Rightarrow\quad\boxed{\angle p \:=\: 55^o}$

In quadrilateral $ABDE,\;\angle q + \angle E + \angle A + \angle p \:=\:360^o$
. . $\angle q + 107^o + 90^o + 55^o \:=\:360^o \quad\Rightarrow\quad \boxed{\angle q \:=\:108^o}$

Since $AE \parallel DI,\;\angle EDI \:=\:\angle DEA \:=\:107^o$

Around vertex $D\!:\;\;\angle r + \angle EDI + \angle q \:=\:360^o$
. . $\angle r + 107^o + 108^o \:=\:360^o \quad\Rightarrow\quad\boxed{\angle r \,=\,145^o}$

Since $\angle DCF + \angle BCF \:=\:180^o,\;\;\angle DCF \,=\,125^o$

So far, we have:
Code:
     A                     B
o - - - - - - - - - o
| 90°          55° *
|                 * 55°          F
|              C o- - - - - - - o
| 107°          * 125°      90° |
E o              *                |
|   *    108° *                 |
|       *    *                t |
|           o  145°             o G
|      107° |                 *
|           |               *
| 90° - 90° | 90°      t  *
o - - - - - o - - - - - o
J            I            H

In hexagon $CFGHID:\;\;2\angle t + 90^o + 145^o + 125^o + 90^o \:=\:720^o$

Therefore: . $2\angle t \,=\,270^o \quad\Rightarrow\quad\boxed{\angle t \:=\:135^o}$

5. Hi buggerzapper,

Am i correct in assuming that the angle p is not 180 degrees?

6. Originally Posted by Archie Meade
Hi buggerzapper,
Am i correct in assuming that the angle p is not 180 degrees?
That was my question also.
If it a straight angle the diagram is certainly misleading.
However, Soroban assumed that it is and he got the same answers as the textbook.

7. Here's another solution, buggerzapper

p=175
r=140
q=113

or

p=170
r=135
q=118

or etc etc etc

there are lots more solutions

why?????

8. Here's a clue...

The angle relationships are...

$q+r=253^o$

$p+q=288^o$

We can find these by drawing triangles inside the shape.

"If" p was 180 degrees, then q would be 108, r would be 145.

If you change r, you can find the changes for the other two angles from it.