+289 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
+289 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
