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?

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- Feb 1st 2010, 02:27 AMsri340triangle 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 AMArchie Meade
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 AMSoroban
Hello, sri340!

Great solution, Archie!

Here's another approach . . .

Quote:

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

The measure of $\displaystyle \angle BAC$ is 42°

. . and segment $\displaystyle ED$ is tangent to the circle at point $\displaystyle C.$

What is the measure of $\displaystyle \angle ACD$ ?

Code:`A`

* o *

* / \ * 138°

* /42°\ *

* / \ *

/ \ o D

* / \ * /

* / \ * /

* / \ */

/ 69° \ /

B o- - - - - - - - -o C

* *

* */

* * * /

/

o E

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

Then $\displaystyle \text{arc}(AC) = 138^o$

. . An inscribed angle is measured by one-half its intercepted arc.

Therefore: .$\displaystyle \angle ACD \,=\,69^o$

. . The angle between a common tangent and secant

. . . . is measured by one-half its intercepted arc.