+0

# Triangles

+1
12
1
+1654

Need help with triangle.  What is the value of x?

Jan 13, 2024

#1
+289
+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

Jan 13, 2024

#1
+289
+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