#1**+1 **

By the Exterior Triangle Angle Sum Theorem ( kinda forgot name) , angle ABC + angle BAC = angle ACD, so

\(x^2-4x-3=9x+27\)

\(x^2-13x-30=0\)

I will use the quadratic formula to find the roots of x, and then eliminate the value of x that produces negative angle measures.

a = 1, b = -13, c = -30

\(x = {13 \pm \sqrt{169+120} \over 2}\)

\(x = {13 \pm 17 \over 2}\)

We get: x = 15, and x = -2.

Eliminate x = -2, and the answer is x = 15

Answer: 15

DS2011 Jan 13, 2024

#1**+1 **

Best Answer

By the Exterior Triangle Angle Sum Theorem ( kinda forgot name) , angle ABC + angle BAC = angle ACD, so

\(x^2-4x-3=9x+27\)

\(x^2-13x-30=0\)

I will use the quadratic formula to find the roots of x, and then eliminate the value of x that produces negative angle measures.

a = 1, b = -13, c = -30

\(x = {13 \pm \sqrt{169+120} \over 2}\)

\(x = {13 \pm 17 \over 2}\)

We get: x = 15, and x = -2.

Eliminate x = -2, and the answer is x = 15

Answer: 15

DS2011 Jan 13, 2024