1. ## tangent circle question

The radius of the incircle of a triangle is 4 cm and the segments in to which one side is divided by the point of contact are 6cm and 8 cm , determine the other two sides of the triangle.
Oh and yeah, R = 4cm and the other R is the name of the contact point..no relation..PLEASE I need help

yeah another tangent Question..

Well, the only way I know to do this Question is by using heron's formula and then comparing the area found by 1/2*b*h,, of the sum of the three triangles that can be formed by joining the centre to the vertices. Is there an easier and less shabby way?

2. Originally Posted by ice_syncer
The radius of the incircle of a triangle is 4 cm and the segments in to which one side is divided by the point of contact are 6cm and 8 cm , determine the other two sides of the triangle.
Oh and yeah, R = 4cm and the other R is the name of the contact point..no relation..PLEASE I need help

yeah another tangent Question..

Well, the only way I know to do this Question is by using heron's formula and then comparing the area found by 1/2*b*h,, of the sum of the three triangles that can be formed by joining the centre to the vertices. Is there an easier and less shabby way?
Apply Pythagoras theorem to triangle ORB. You know OR and BR. So you'll find OB.
Then apply Pythagoras theorem to triangle OPB. You know OP and OB. So you'll find BP.
And so on...

I don't know if it's quicker than applying Heron's formula.

3. $\displaystyle \frac\beta{2}=\arctan\frac{4}{8}$ ($\displaystyle \triangle ORB$)
$\displaystyle \frac\gamma{2}=\arctan\frac{4}{6}$ ($\displaystyle \triangle ORC$)

$\displaystyle \Downarrow$

$\displaystyle \alpha=180°-\beta-\gamma$

$\displaystyle \frac{8+6}{\sin A} = \frac{b}{\sin\beta}$ (Law of sines; same for c)