# Help! 2 questions - triangles and circles

• Nov 1st 2008, 05:02 AM
BG5965
Help! 2 questions - triangles and circles

1) In the figure, two circles touch each other at C. AB is a common tangent touching the circles at A and B and angle ABC is 35 degrees. Find the angle CAB.
http://i301.photobucket.com/albums/nn74/BL5965/8.jpg

2) In the figure, O is the centre of the circle. A and B are two points on the circle so that triangle OAB is an equilateral triangle. OA is produced to C so that OA = AC. Find the angle ABC and prove whether CB is tangent to the circle at B, giving a reason.
http://i301.photobucket.com/albums/nn74/BL5965/9.jpg

Thankyou so much for any help, I don't know where to start in solving these.
• Nov 1st 2008, 10:48 AM
Soroban
Hello, BG5965!

Quote:

2) In the figure, O is the centre of the circle.
A and B are two points on the circle so that triangle OAB is an equilateral triangle.
OA is produced to C so that OA = AC.
Find the angle ABC and prove whether CB is tangent to the circle at B.
http://i301.photobucket.com/albums/nn74/BL5965/9.jpg

Let $\displaystyle r \;=\;AO = BO$, radius $\displaystyle r.$
. . Then: .$\displaystyle CO = 2r$

In $\displaystyle \Delta COB,\;\angle COB = 60^o,\;BO = r,\;CO = 2r$

Then: .$\displaystyle \sin(\angle C) \:=\:\frac{r}{2r} \:=\:\frac{1}{2} \quad\Rightarrow\quad \angle C \:=\:30^o$

Hence: .$\displaystyle \angle OBC = 90^o$ and $\displaystyle OB$ is tangent to the circle.

Since $\displaystyle \angle OBA = 60^o$, then $\displaystyle \angle ABC = 30^o$