# triangle inscribed in the circle

• Feb 1st 2010, 02:27 AM
sri340
triangle inscribed in the circle
triangle ABC is inscribed in the circle and AC = AB. The measure of angle BAC is 42 degrees and segment ED is tangent to the circle at point C. What is the measure of angle ACD?
• Feb 1st 2010, 03:21 AM
Quote:

Originally Posted by sri340
triangle ABC is inscribed in the circle and AC = AB. The measure of angle BAC is 42 degrees and segment ED is tangent to the circle at point C. What is the measure of angle ACD?

If you draw a circle, locate the centre, then the inscribed triangle is isosceles.
Join the centre to all 3 triangle vertices.
The angle at A is now split into 2 equal 21 degree angles.
If we label the centre F, then angle FAB = angle FAC = 21 degrees.
The 3 triangles within the inscribed triangle ABC are all isosceles also.
Hence, angle FCA = angle FBA = 21 degrees.

The tangent is perpendicular to the line CF.
Therefore the sum of angles FCA and ACD is 90 degrees.
Hence angle ACD is 90-21 = 69 degrees.
• Feb 1st 2010, 06:38 AM
Soroban
Hello, sri340!

Great solution, Archie!

Here's another approach . . .

Quote:

Triangle $ABC$ is inscribed in the circle and $AC = AB.$

The measure of $\angle BAC$ is 42°
. . and segment $ED$ is tangent to the circle at point $C.$

What is the measure of $\angle ACD$ ?

Code:

                A               * o *           *    / \    *  138°         *    /42°\    *       *    /    \    *             /      \        o D       *    /        \    *  /       *  /          \  * /       *  /            \  */         / 69°          \ /     B o- - - - - - - - -o C         *              *           *          */               * * *  /                     /                     o E

Since $\Delta ABC$ is isosceles, $\angle B = \angle C = 69^o$

Then $\text{arc}(AC) = 138^o$
. .
An inscribed angle is measured by one-half its intercepted arc.

Therefore: . $\angle ACD \,=\,69^o$
. .
The angle between a common tangent and secant
. . . . is measured by one-half its intercepted arc.