Circle and Triangle

• Oct 1st 2008, 04:56 AM
shaurya
Circle and Triangle
In the given figure i need to find AC and AB.
CD= 6 cm
BD = 8 cm

Cant figure this thing out.
• Oct 1st 2008, 07:17 AM
masters
Quote:

Originally Posted by shaurya
In the given figure i need to find AC and AB.
CD= 6 cm
BD = 8 cm

Cant figure this thing out.

Can we assume that $\displaystyle \overline{OD}$ is perpendicular to $\displaystyle \overline{CB}$ at the point of tangency D?

Draw OX and OY perpendicular to AC and AB, respectively, at their points of tangency.

Center O is the incenter of the inscribed circle found by the intersection of the three angle bisectors. Thus,

Angle ACO = Angle BCO
Angle CAO = Angle BAO
Angle ABO = Angle CBO

CD = CX and BD = BY, since tangents to a circle from an external point are equal.

CX = 6

BY = 8

Use Arctan to find angles OCD and OBD.

Angle OCD = $\displaystyle \arctan \frac{4}{6} \approx 33.69^{\circ}$

Angle OBD = $\displaystyle \arctan \frac{4}{8} \approx 26.57^{\circ}$

Since the angles at C and B were bisected,

Angle ACB = 2(33.69) = 67.38
Angle ABC = 2(26.57) = 53.13

This makes Angle CAB = 180 - 67.38 - 53.13 = 59.49

Half that gives Angle CAD = 29.75

Use Tangent to find AX.

$\displaystyle \tan 29.75=\frac{4}{AX}$

AX = AY because they are tangent segments from the same external point.

$\displaystyle AX \approx 7$
$\displaystyle AY \approx 7$

Therefore, AC = 13 and AB = 15, approximately.
• Oct 1st 2008, 09:02 AM
shaurya
well, i coudnt quite understand the trignometric term arctan and couldnt understand how it became 7 in last.
Maybe you havce used a log table but we arent given log tables and all.
The answer is right. We did it long ago in class using herons formula or something.
• Oct 1st 2008, 09:10 AM
masters
Quote:

Originally Posted by shaurya
well, i coudnt quite understand the trignometric term arctan and couldnt understand how it became 7 in last.
Maybe you havce used a log table but we arent given log tables and all.
The answer is right. We did it long ago in class using herons formula or something.

Arctan is the same as $\displaystyle \tan^{-1}$. Use this to find the measure of an angle when you know the tangent ratio.

Heron's formula is used to find the area of a triangle if only the sides are known. I'm not sure how that would fit in this situation unless the question were ammended to find the area after we find AB and AC.
• Oct 1st 2008, 09:17 AM
masters
Quote:

Originally Posted by shaurya
well, i coudnt quite understand the trignometric term arctan and couldnt understand how it became 7 in last.

$\displaystyle \tan 29.75=\frac{4}{AX}$

$\displaystyle (AX)\tan 29.75=4$

$\displaystyle AX=\frac{4}{\tan 29.75}$

$\displaystyle AX \approx 6.999 \approx 7$