drawing a blank (1 Viewer)

[fashionista]

woopsies...sorry :$

 

[fashionista]

thanks for ur help!!!!!!!!!!!!!!!!!!!!!

 

[fashionista]

k..i reeeely hope this is the last time i need help with these questions i really really do
um..this one right here is givin me a bit of the troublings
(its a different set of questions)

25. C is any point on a circle diameter AB. P and Q are points on the minor arcs AC and BC. Prove that < APC + < CQB = 3 right angles (270 degrees)

 

[Xayma]

Originally posted by fashionista
25. C is any point on a circle diameter AB. P and Q are points on the minor arcs AC and BC. Prove that < APC + < CQB = 3 right angles (270 degrees)

Let < APC=a
therefore < ABC=[180-a] degrees (cyclic quadrilateral)

Let < CQB=b
therefore < CAB=[180-b] degrees (cyclic quadrilateral)

therefore < ACB=[-180+a+b]degrees (Angle sum of triangle)
but < ACB=90degrees (angle in semicircle is a right angle)
therefore 90 degrees=(-180+a+b)degrees
therefore a+b=270 degrees
therefore < APC + < CQB=270 degrees

 

[fashionista]

TANKING U MUCHLY

 

[Xayma]

Originally posted by fashionista
TANKING U MUCHLY

No problems.

 

[fashionista]

heres another one how sad.....ABCD is a cyclic quadrilateral and FAE is a tnagent at A. Angle DAE = 50 degress. Calculate the size of < BAF

 

[Xayma]

was there a picture with it, because naming the tangent FAE seems to imply that < DAE will be in part of the cyclic quad.

 

[fashionista]

yes ther was..ill try n stik it in for ya

 

[Xayma]

Thanks

 

[fashionista]

this is gay..its too big..
u draw a circle and inscribe the quadrilateral ABCD...then at A u draw a tangent and the end of the tangent closest to B is F and the end of the tangent closest to D is E.


i hope that made some sense

 

[Xayma]

Yeah thats what I thought it would be but calling it FAE confused me.

 

[Xayma]

There must be more information than whats in your post becaue I decided to draw it cause I couldnt work out what properties I use depending on where you draw it you will get different angles.

 

[fashionista]

which post are u talkin bout..the one with the question ot the 'descriptive' one? cuz the question is pretty much copied from the book...if u have fitzpatrick 3u its question 20 on page 17

 

[Xayma]

Originally posted by fashionista
heres another one how sad.....ABCD is a cyclic quadrilateral and FAE is a tnagent at A. Angle DAE = 50 degress. Calculate the size of < BAF

This one, < BAF can be multiple sizes, even if DAE=50 degrees.

 

[KeypadSDM]

Originally posted by Xayma
This one, < BAF can be multiple sizes, even if DAE=50 degrees.

Because, althought D is stagnant, there are no limitations on the position of B and C

Hence the chord BD can change length, and thus create different sized angles: 130 > (< DAB) > 0

 

[Xayma]

Originally posted by KeypadSDM
Because, althought D is stagnant, there are no limitations on the position of B and C

Hence the chord BD can change length, and thus create different sized angles: 130 > (< DAB) > 0

Yeah exactly what I was trying to say. I had a little diagram as well but it didnt scan properly

 

[KeypadSDM]

Originally posted by Xayma
Yeah exactly what I was trying to say. I had a little diagram as well but it didnt scan properly

Do them in paint, it's pretty effective, and you can make circles easily

I think the question probably had a set of paralell lines, but where, I don't know.