Thread: Trigonometry Problem help please

I am really stuck on a trigonometry problem please help if you can.
---------------------------
A circle has a radius of 'r'
And its center is at 'C'
The distance from 'A' to 'B' is 'x'
find 'r' if
A = 45 degrees and 'x' equals 15

C
-|-\
-|---\
r|-----\r< C to B is r and C to D is also r
-|-------\
-|---------\B< B to A is 15
-|-----------\
-|_____45\x < x equals 15 and 45 is Angle A
D-------------A

Thats how it is put in the book and The answer is 36 but I do not know how to solve this
Because r ends up on both sides of the equation.
if I go
sin45 = r/r+15 I get
(r+15)(sin45) = r
and so on I really could use some help here because I need trig for programming class. thanks

Hello ryogaki.

Here's what I think the diagram looks like:




You do not need to use the sine rule because it is a 45 degree triangle. Since a length is tangent to the circle and another goes through the centre, we know that it is a right angle triangle, and so we can use the pythagorus theorum.

Length CD = AD because it is a 45 degree triangle.

Length BC also equals CD and AD because it is the radius of the same circle.

Let BC = CD = AD = x

Now using pythagorus theorum:

We know that length AC = (x + 15)

Therefore the equation becomes:

Now can you carry on from there?

well heres what the picture really looks like.
I see what your saying about the 45 degree angle but r is one side and only a part of the other. The triangle comes out further on the CA side
Also if I just do that equation C squared = a squared + b squared it does not give a value for r and the book says that the answer should be 'r = 36'

Have you tried solving the equation of ?

In my instance, x = r. So solve for x and you should get your radius.

EDIT: Howcome the CA side comes out further? A 45 degree right angle triangle has 2 equal lengths.

Thats what I thought too... it should work out like a 45 45 90 triangle.
The problem is r is a side but then theres an r + 15 so AB can not just be the radius.
and every way I try to solve this gives me r on both sides...
I used your way and csc45 and (t) (t) (t times the square root of 2)
sec of angle C I dont get how to get r to equal 36...

r^2 + r^2 = (r + 15)^2

2r^2 = ( r+15)^2

sqrt(2)*r = r+15

r[sqrt(2) - 1] = 15

r = 15/[sqrt(2) - 1]

Now solve for r.

i do not understand...in step 3 you have r on both sides but then in step four the right r is gone. how did that work?

How sa-ri-ga-ma got from step 3 to step 4:



(Subtract r from bith sides)



(Factorise the r)

OMG! I have been so messed up on this whole problem for the littlest of reasons...
I was subtracting r from r [squareroot 2] instead of factoring....that solution does work and it worked on all the other questions like that. I can not believe after all that algebra I would miss something so simple... Thank you Educated and sa-ri-ga-ma very much. That little thing had me going crazy. stick around I might need help with 1+1 next..lol!